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<blockquote data-quote="copemuadong_dn9" data-source="post: 72431" data-attributes="member: 74949">[FONT=.VnTime]Bµi 1. [/FONT][FONT=.VnTime]&nbsp; &nbsp; Hçn hîp X gåm 2 kim lo¹i Al vµ Cu. Cho 18,2 gam X vµo 100ml dung dÞch Y chøa H2SO4 12M vµ HNO3 2M, ®un nãng t¹o ra dung dÞch Z vµ 8,96 lÝt (®ktc) hçn hîp T gåm NO vµ khÝ D kh«ng mµu. Hçn hîp T cã tû khèi so víi hidro = 23,5. [/FONT][FONT=.VnTime]TÝnh khèi l&amp;shy;îng mçi kim lo¹i trong hçn hîp ®Çu vµ l&amp;shy;îng mçi muèi trong dung dÞch Z[/FONT][FONT=.VnTime] [/FONT][FONT=.VnArialH]Gi¶i:[/FONT][FONT=.VnArialH] [/FONT][FONT=.VnTime] M[/FONT][FONT=.VnTime]T =23,5 . 2 = 47 [/FONT]®[FONT=.VnTime] M[/FONT][FONT=.VnTime]NO = 30 &lt; 47 &lt; MD.&nbsp; [/FONT]®[FONT=.VnTime] D lµ SO[/FONT][FONT=.VnTime]2 = 64[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Suy ra sè mol NO = 0,2 mol vµ SO2 = 0,2 mol[/FONT][FONT=.VnTime]Thùc chÊt ph¶n øng theo c¸c ph&amp;shy;¬ng tr×nh sau:[/FONT][FONT=.VnTime]&nbsp;&nbsp; Al – 3e [/FONT]®[FONT=.VnTime] Al3+.&nbsp; Víi sè mol Al = x vµ sè mol Cu = y[/FONT][FONT=.VnTime]&nbsp;&nbsp; Cu – 2e [/FONT]®[FONT=.VnTime] Cu2+. Tæng sè e nh&amp;shy;êng = 3x + 2y[/FONT][FONT=.VnTime]&nbsp; &nbsp; NO3- + 4H+ + 3e [/FONT]®[FONT=.VnTime] NO + 2H2O [/FONT][FONT=.VnTime]&nbsp; &nbsp;&nbsp; SO42– + 4H+ + 2e [/FONT]®[FONT=.VnTime] SO2 + 2H2O&nbsp;&nbsp; (Tæng sè mol e thu = 0,6 + 0,4 = 1 )[/FONT][FONT=.VnTime]Sè mol H+ trong Y = 1,2 . 2 + 0,2 = 2,6&nbsp; &gt;&nbsp; sè mol H+dù ph¶n øng = (0,2 + 0,2) 4= 1,6 [/FONT][FONT=.VnTime][/FONT][FONT=.VnTime]Nªn kim lo¹i tan hÕt.[/FONT][FONT=.VnTime]VËy ta cã hÖ ph&amp;shy;¬ng tr×nh:&nbsp; &nbsp;&nbsp; 27x + 64y = 18,2[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 3x + 2y&nbsp;&nbsp; =&nbsp; 1&nbsp; [/FONT]®[FONT=.VnTime] gi¶i pt cho x = y = 0,2[/FONT][FONT=.VnTime]V× NO3– ph¶n øng =&nbsp; NO3– trong Y nªn dung dÞch Z kh«ng cã NO3–‑ vµ chØ cã Al3+, Cu2+, SO42–. L&amp;shy;îng Al2(SO4)3 = . 342 = 34,2 gam[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; L&amp;shy;îng CuSO4 = 0,2 . 160 = 32 gam[/FONT][FONT=.VnTime]Bµi 2. [/FONT][FONT=.VnTime]&nbsp; Hßa tan a gam hçn hîp Cu vµ Fe (trong ®ã Fe chiÕm 30% vÒ khèi l&amp;shy;îng) b»ng 50 ml dung dÞch HNO3 63% (d = 1,38 g/ml), khuÊy ®Òu cho tíi ph¶n øng hoµn toµn thu ®&amp;shy;îc chÊt r¾n A c©n nÆng 0,75a gam , dung dÞch B vµ 6,104 lÝt hçn hîp khÝ NO vµ NO2 (®ktc). Hái c« c¹n dung dÞch B thu ®&amp;shy;îc bao nhiªu gam muèi khan?[/FONT][FONT=.VnArialH]Gi¶i:[/FONT][FONT=.VnArialH]&nbsp;&nbsp; [/FONT][FONT=.VnTime]PhÇn hçn hîp tan = 0,25a gam &lt; 0,3a gam nªn Fe cßn d&amp;shy; vµ Cu ch&amp;shy;a ph¶n øng[/FONT][FONT=.VnTime]&nbsp; &nbsp;&nbsp; C¸c ph¶n øng:&nbsp; &nbsp; &nbsp; &nbsp; Fe + 4 HNO3 [/FONT]®[FONT=.VnTime] Fe(NO3)3 + NO [/FONT]&amp;shy;[FONT=.VnTime]+ 2H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; Fe + 6 HNO3 [/FONT]®[FONT=.VnTime]&nbsp; Fe(NO3)3 + 3NO[/FONT][FONT=.VnTime]2 [/FONT]&amp;shy;[FONT=.VnTime]+ 3H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp;&nbsp; Do Fe d&amp;shy; nªn cßn c¸c ph¶n øng: Fe +&nbsp; 2Fe(NO3)3&nbsp; [/FONT]®[FONT=.VnTime]&nbsp; 3Fe(NO3)2 [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Tøc lµ trong dung dÞch B chØ cã Fe(NO3)2[/FONT][FONT=.VnTime]&nbsp; &nbsp;&nbsp; Vµ l&amp;shy;îng muèi = l&amp;shy;îng Fe + l&amp;shy;îng NO3– [/FONT][FONT=.VnTime]&nbsp; &nbsp;&nbsp; Mµ sè mol NO3– cßn l¹i trong B = tæng sè mol NO3–&nbsp; – sè mol khÝ [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; = 0,69 – 0,2725 = 0,4175 mol[/FONT][FONT=.VnTime]&nbsp; &nbsp; VËy l&amp;shy;îng muèi khan = . 180 = 37,575 gam[/FONT][FONT=.VnTime]Bµi 3.&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; Hßa tan 17,4 gam hçn hîp 3 kim lo¹i Al, Cu, Fe trong dung dÞch HCl d&amp;shy; thÊy tho¸t ra 8,96 lÝt khÝ(®ktc). NÕu cho 34,8 gam hçn hîp trªn t¸c dông víi dung dÞch CuSO4 d&amp;shy; råi läc chÊt r¾n t¹o ra hßa tan b»ng HNO3 th× tho¸t ra 26,88 lÝt khÝ (®ktc) cã tû khèi so víi oxi = 1,27. ViÕt c¸c ph&amp;shy;¬ng tr×nh ph¶n øng vµ tÝnh thµnh phÇn hçn hîp ban ®Çu. [/FONT][FONT=.VnArialH]Gi¶i:&nbsp; &nbsp; [/FONT][FONT=.VnTime]2Al + 6 HCl [/FONT]®[FONT=.VnTime] 2AlCl3 + 3H[/FONT][FONT=.VnTime]2[/FONT]&amp;shy;[FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (1)[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Fe + 2 HCl [/FONT]®[FONT=.VnTime]&nbsp; FeCl2&nbsp; + H[/FONT][FONT=.VnTime]2[/FONT]&amp;shy;[FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; (2)[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2Al + 3CuSO4 [/FONT]®[FONT=.VnTime] Al2(SO4)3 + 3Cu[/FONT]¯[FONT=.VnTime]&nbsp; &nbsp;&nbsp; (3) [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Fe + CuSO4 [/FONT]®[FONT=.VnTime] FeSO4 + Cu[/FONT]¯[FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; (4)[/FONT][FONT=.VnTime]KhÝ cã&nbsp;&nbsp; 30 &lt; MTB = 40,64 &lt;&nbsp; 46&nbsp; lµ hçn hîp&nbsp; NO vµ NO2 (cÇn chó ý lµ trong c¸c s¶n phÈm chøa N t¹o ra khi t¸c dông víi HNO3 ngoµi muèi, Cu chØ t¹o NO vµ NO2 )[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; Cu + 4HNO3 [/FONT]®[FONT=.VnTime] Cu(NO3)2 + 2NO[/FONT][FONT=.VnTime]2 [/FONT]&amp;shy;[FONT=.VnTime]+ 2H2O&nbsp; &nbsp; &nbsp;&nbsp; (5)[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 3Cu + 8HNO3 [/FONT]®[FONT=.VnTime] 3Cu(NO3)2 + 2NO [/FONT]&amp;shy;[FONT=.VnTime]+ 4H2O&nbsp; &nbsp; &nbsp;&nbsp; (6)[/FONT][FONT=.VnTime]Tõ MTB = 40,64 vµ tæng sè mol = 1,2 tÝnh ®&amp;shy;îc NO = 0,4 mol vµ NO2= 0,8 mol[/FONT][FONT=.VnTime]Theo pt (5), (6) Cu = 0,4 + 0,6 = 1,0 mol[/FONT][FONT=.VnTime]So s¸nh (1), (2) víi (3), (4) thÊy sè mol Cu[/FONT]¯[FONT=.VnTime] = H2[/FONT]&amp;shy;[FONT=.VnTime]= 0,8 mol (TN 2 lÊy l&amp;shy;îng gÊp ®«i)[/FONT][FONT=.VnTime]Suy ra Cu ban ®Çu = = 0,1 mol [/FONT]~[FONT=.VnTime] 6,4 gam [/FONT]®[FONT=.VnTime]&nbsp; Al + Fe = 11 gam[/FONT][FONT=.VnTime]Tõ hÖ pt vÒ tæng l&amp;shy;îng Al + Fe vµ tæng sè mol H2[/FONT]&amp;shy;[FONT=.VnTime] tÝnh ®&amp;shy;îc Al = 5,4 g vµ Fe = 5,6 g[/FONT][FONT=.VnTime]Bµi 4.&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; Trén CuO víi mét oxit kim lo¹i ®¬n hãa trÞ II theo tû lÖ mol 1:2 ®&amp;shy;îc hçn hîp A. DÉn mét luång khÝ H2 d&amp;shy; ®i qua 3,6 gam A nung nãng thu ®&amp;shy;îc hçn hîp B. §Ó hßa tan hÕt B cÇn 60 ml dung dÞch HNO3 nång ®é 2,5M vµ thu ®&amp;shy;îc V lÝt khÝ NO duy nhÊt(®ktc) vµ dung dÞch chØ chøa nitat kimlo¹i. X¸c ®Þnh kim lo¹i hãa trÞ II nãi trªn vµ tÝnh V.[/FONT][FONT=.VnTime] [/FONT][FONT=.VnArialH]Gi¶i: [/FONT][FONT=.VnTime]Gäi oxit kim lo¹i ph¶i t×m lµ MO vµ a vµ 2a lµ sè mol CuO vµ MO trong A.[/FONT][FONT=.VnTime]V× hidro chØ khö ®&amp;shy;îc nh÷ng oxit kim lo¹i ®øng sau nh«m trong d·y ®iÖn hãa nªn cã 2 kh¶ n¨ng x¶y ra:[/FONT][FONT=.VnTime]* Tr&amp;shy;êng hîp 1: M ®øng sau nh«m trong d·y ®iÖn hãa[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; CuO&nbsp; &nbsp; +&nbsp; &nbsp; H2&nbsp; &nbsp; [/FONT]®[FONT=.VnTime]&nbsp; &nbsp; Cu&nbsp; &nbsp; +&nbsp; &nbsp; H2O&nbsp;&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; MO&nbsp; &nbsp;&nbsp; +&nbsp; &nbsp; H2&nbsp; &nbsp; [/FONT]®[FONT=.VnTime]&nbsp; &nbsp; M&nbsp; &nbsp; +&nbsp; &nbsp; H2O&nbsp;&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 3Cu + 8HNO3 [/FONT]®[FONT=.VnTime] 3Cu(NO3)2 + 2NO [/FONT]&amp;shy;[FONT=.VnTime]+ 4H2O&nbsp; &nbsp; &nbsp;&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 3M&nbsp; + 8HNO3 [/FONT]®[FONT=.VnTime] 3 M(NO3)2 + 2NO [/FONT]&amp;shy;[FONT=.VnTime]+ 4H2O&nbsp; &nbsp;&nbsp; [/FONT][FONT=.VnTime]Ta cã hÖ pt:&nbsp; &nbsp; Gi¶i hÖ pt cho a = 0,01875 vµ M = 40 [/FONT]~[FONT=.VnTime] Ca[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp;&nbsp; Tr&amp;shy;êng hîp nµy lo¹i v× Ca ®øng tr&amp;shy;íc Al trong d·y thÕ ®iÖn hãa.[/FONT][FONT=.VnTime]* Tr&amp;shy;êng hîp 2: M ®øng tr&amp;shy;íc nh«m trong d·y ®iÖn hãa[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; CuO&nbsp; &nbsp; +&nbsp; &nbsp; H2&nbsp; &nbsp; [/FONT]®[FONT=.VnTime]&nbsp; &nbsp; Cu&nbsp; &nbsp; +&nbsp; &nbsp; H2O&nbsp;&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 3Cu + 8HNO3 [/FONT]®[FONT=.VnTime] 3Cu(NO3)2 + 2NO [/FONT]&amp;shy;[FONT=.VnTime]+ 4H2O&nbsp; &nbsp; &nbsp;&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; MO&nbsp; + 2HNO3 [/FONT]®[FONT=.VnTime]&nbsp; M(NO3)2 + 2H2O [/FONT][FONT=.VnTime]Ta cã hÖ pt:&nbsp; &nbsp;&nbsp; Gi¶i hÖ pt cho a = 0,01875 vµ M = 24 [/FONT]~[FONT=.VnTime] Mg[/FONT][FONT=.VnTime]NghiÖm nµy hîp lý vµ V= . 22,4 = 0,28 lÝt. [/FONT]®[FONT=.VnTime] [/FONT][FONT=.VnTime]Bµi 5.&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; X,Y lµ kim lo¹i ®¬n hãa trÞ II vµ III . Hßa tan hÕt 14,0 gam hçn hîp X,Y b»ng axit HNO3 tho¸t ra 14,784 lÝt (27,30C vµ 1,1atm) hçn hîp 2 khÝ oxit cã mµu n©u vµ cã tû khèi so víi He = 9,56 , dung dÞch nhËn ®&amp;shy;îc chØ chøa nitrat kim lo¹i. Cïng l&amp;shy;îng hçn hîp 2 kim lo¹i trªn cho t¸c dông víi axit HCl d&amp;shy; th× còng tho¸t ra 14,784 lÝt khÝ (27,30C vµ 1atm) vµ cßn l¹i 3,2 gam chÊt r¾n kh«ng tan.[/FONT][FONT=.VnTime]X¸c ®Þnh X, Y vµ tÝnh % l&amp;shy;îng mçi kim lo¹i trong hçn hîp ®Çu.[/FONT][FONT=.VnArialH]Gi¶i: [/FONT][FONT=.VnTime]Sè mol khÝ = 0,66 vµ 0,6.&nbsp; [/FONT][FONT=.VnTime] Tõ MTB= 9,56. 4 = 38,24 suy ra NO2 &gt; 38,24 nªn khÝ cßn l¹i ph¶i lµ NO = 30 &lt; 38,24[/FONT][FONT=.VnTime]&nbsp; Vµ tÝnh ®&amp;shy;îc NO = 0,32 mol vµ NO2 = 0,34 mol[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 3X + 8HNO3 [/FONT]®[FONT=.VnTime] 3X(NO3)2 + 2NO [/FONT]&amp;shy;[FONT=.VnTime]+ 4H2O&nbsp; &nbsp; &nbsp;&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Y&nbsp; + 4HNO3 [/FONT]®[FONT=.VnTime]&nbsp; Y(NO3)3&nbsp;&nbsp; +&nbsp; NO [/FONT]&amp;shy;[FONT=.VnTime]+ 2H2O&nbsp; &nbsp;&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; X + 4HNO3 [/FONT]®[FONT=.VnTime]&nbsp; X(NO3)2&nbsp; + 2NO[/FONT][FONT=.VnTime]2 [/FONT]&amp;shy;[FONT=.VnTime]+ 2H2O&nbsp; &nbsp; &nbsp;&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; Y + 6HNO3 [/FONT]®[FONT=.VnTime]&nbsp; Y(NO3)3&nbsp; + 3NO[/FONT][FONT=.VnTime]2 [/FONT]&amp;shy;[FONT=.VnTime]+ 3H2O [/FONT][FONT=.VnTime]&nbsp; &nbsp; X + 2HCl [/FONT]®[FONT=.VnTime] XCl2 + H[/FONT][FONT=.VnTime]2 [/FONT]&amp;shy;[FONT=.VnTime]&nbsp;&nbsp; hoÆc&nbsp;&nbsp; 2Y + 6HCl [/FONT]®[FONT=.VnTime] 2YCl3 + 3H[/FONT][FONT=.VnTime]2 [/FONT]&amp;shy;[FONT=.VnTime] [/FONT][FONT=.VnTime]BiÖn luËn: * NÕu kim lo¹i Y kh«ng tan trong axit HCl[/FONT][FONT=.VnTime]&nbsp;&nbsp; Theo pt: sè mol X = 0,6 vµ l&amp;shy;îng X = 10,8 gam nªn&nbsp; X = = 18 (kh«ng tháa m·n kim lo¹i nµo)[/FONT][FONT=.VnTime]* VËy kim lo¹i X kh«ng tan trong axit HCl [/FONT][FONT=.VnTime]&nbsp;&nbsp; Theo pt: sè mol Y = 0,4 vµ l&amp;shy;îng Y = 14- 3,2= 10,8 gam nªn Y = = 27 [/FONT]~[FONT=.VnTime] Al [/FONT][FONT=.VnTime]§Æt sè mol X b»ng a:&nbsp; &nbsp;&nbsp; tæng sè e nh&amp;shy;êng = 0,4. 3 + 2a = 1,2 + 2a [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; tæng sè e thu = 0,32. 3 + 0,34 = 1,30&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; Theo qui t¾c b¶o toµn sè mol e: 1,2 + 2a = 1,3 [/FONT]®[FONT=.VnTime] a = 0,05[/FONT][FONT=.VnTime]&nbsp; &nbsp; VËy X = = 64 [/FONT]~[FONT=.VnTime] Cu&nbsp;&nbsp; vµ&nbsp; % Al = 77,14%&nbsp; ;&nbsp; %Cu = 22,86%[/FONT][FONT=.VnTime]Bµi 6.&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; Cho 21,52 gam hçn hîp A gåm kim lo¹i X ®¬n hãa trÞ II vµ muèi nitrat cña nã vµo b×nh kÝn dung tÝch kh«ng ®æi 3 lÝt (kh«ng chøa kh«ng khÝ) råi nung b×nh ®Õn nhiÖt ®é cao ®Ó ph¶n øng x¶y ra hoµn toµn, s¶n phÈm thu ®&amp;shy;îc lµ oxit kim lo¹i. Sau ph¶n øng ®&amp;shy;a b×nh vÒ 54,60C th× ¸p suÊt trong b×nh lµ P. Chia ®«i chÊt r¾n trong b×nh sau ph¶n øng: phÇn 1 ph¶n øng võa ®ñ víi 667ml dung dÞch HNO3 nång ®é 0,38M tho¸t ra khÝ NO duy nhÊt vµ dung dÞch chØ chøa nitrat kim lo¹i. PhÇn 2 ph¶n øng võa hÕt víi 300ml dung dÞch H2SO4 lo·ng 0,2M thu ®&amp;shy;îc dung dÞch B.[/FONT][FONT=.VnTime]&nbsp; &nbsp; a) X¸c ®Þnh kim lo¹i X vµ tÝnh % l&amp;shy;îng mçi chÊt trong A.[/FONT][FONT=.VnTime]&nbsp; &nbsp; b) TÝnh P.[/FONT][FONT=.VnArialH]Gi¶i:&nbsp; [/FONT][FONT=.VnTime]sè mol HNO3 = 0,38. 0,667 = 0,25346&nbsp; vµ sè mol H2SO4 = 0,3. 0,2 = 0,06&nbsp; [/FONT][FONT=.VnArialH] [/FONT][FONT=.VnTime]§Æt sè mol X(NO3)2 vµ X ban ®Çu lµ a vµ b.[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2X(NO3)2&nbsp;&nbsp; [/FONT]®[FONT=.VnTime] 2XO + 4NO2 + O[/FONT][FONT=.VnTime]2 [/FONT]&amp;shy;[FONT=.VnTime]&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; a&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; a&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2a&nbsp; &nbsp; &nbsp; 0,5a[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 2X + O2 [/FONT]®[FONT=.VnTime] 2XO&nbsp;&nbsp; do ph¶n øng víi HNO3 cã khÝ NO[/FONT]&amp;shy;[FONT=.VnTime] nªn X cã d&amp;shy; vµ [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; a&nbsp; &nbsp;&nbsp; 0,5a&nbsp; &nbsp; &nbsp;&nbsp; a&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; phÇn d&amp;shy; = b – a (mol)[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; XO + 2HNO3 [/FONT]®[FONT=.VnTime] X(NO3)2 + H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 3X + 8HNO3 [/FONT]®[FONT=.VnTime] 3X(NO3)2 + 2NO [/FONT]&amp;shy;[FONT=.VnTime]+ 4H2O [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; XO + H2SO4 [/FONT]®[FONT=.VnTime] XSO4 + H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; X&nbsp; &nbsp; + H2SO4 [/FONT]®[FONT=.VnTime] XSO4 + H[/FONT][FONT=.VnTime]2[/FONT]&amp;shy;[FONT=.VnTime]&nbsp; Theo pt: sè mol HNO3 ph¶n øng = 2a + (b - a) =&nbsp; 0,253&nbsp;&nbsp; hay&nbsp;&nbsp; a + 2b = 0,38&nbsp; (1)[/FONT][FONT=.VnTime]BiÖn luËn: * NÕu M ®øng tr&amp;shy;íc hidro trong d·y ®iÖn hãa th× theo pt[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp;&nbsp; a +&nbsp; (b - a) = 0,06&nbsp; hay a + b = 0,12&nbsp;&nbsp; (2). Gi¶i (1)(2) cho a = – 0,14 &lt; 0 (lo¹i)[/FONT][FONT=.VnTime]* VËy M ®øng sau hidro trong d·y ®iÖn hãa vµ kh«ng t¸c dông víi H2SO4 lo·ng,[/FONT][FONT=.VnTime]&nbsp; &nbsp; khi ®ã a = 0,06 [/FONT]®[FONT=.VnTime] b = 0,16 vµ 0,06(M + 124) + 0,16M = 21,52 [/FONT]®[FONT=.VnTime] M = 64 [/FONT]~[FONT=.VnTime] Cu[/FONT][FONT=.VnTime]Suy ra % Cu = 47,5 % vµ % Cu(NO3)2 = 52,5%[/FONT][FONT=.VnTime]&nbsp; Sau khi nung trong b×nh chøa 0,12 mol NO2 nªn P = = 1,07 atm.[/FONT][FONT=.VnTime]Bµi 7.&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; Hßa tan hoµn toµn 2,52 gam hçn hîp Mg vµ Al b»ng dung dÞch HCl thu ®&amp;shy;îc 2,688 lÝt hidro (®ktc). Còng l&amp;shy;îng hçn hîp nµy nÕu hßa tan hoµn toµn b»ng H2SO4 ®Æc nãng th× thu ®&amp;shy;îc 0,03 mol mét s¶n phÈm duy nhÊt h×nh thµnh do sù khö S+6. [/FONT][FONT=.VnTime]&nbsp; &nbsp;&nbsp; a) X¸c ®Þnh s¶n phÈm duy nhÊt nãi trªn.[/FONT][FONT=.VnTime]&nbsp; &nbsp;&nbsp; b) NÕu hßa tan hoµn toµn còng l&amp;shy;îng hçn hîp trªn b»ng dung dÞch HNO3 10,5%[/FONT][FONT=.VnTime](d =1,2 g/ml) th× thu ®&amp;shy;îc 0,03 mol mét s¶n phÈm duy nhÊt h×nh thµnh do sù khö N+5.[/FONT][FONT=.VnTime]TÝnh thÓ tÝch tèi thiÓu dung dÞch HNO3 ®· dïng.[/FONT][FONT=.VnArialH]Gi¶i: [/FONT][FONT=.VnTime]a)[/FONT][FONT=.VnArialH]&nbsp; &nbsp; &nbsp; [/FONT][FONT=.VnTime]Mg + 2HCl [/FONT]®[FONT=.VnTime] MgCl2 + H[/FONT][FONT=.VnTime]2[/FONT]&amp;shy;[FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2Al + 6HCl [/FONT]®[FONT=.VnTime] 2AlCl3 + 3H[/FONT][FONT=.VnTime]2[/FONT]&amp;shy;[FONT=.VnTime]LËp hÖ pt vµ gi¶i hÖ pt cho Mg = 0,06 mol vµ Al = 0,04 mol[/FONT][FONT=.VnArialH]&nbsp; &nbsp; [/FONT][FONT=.VnTime]§Æt sè oxihãa cña l&amp;shy;u huúnh trong s¶n phÈm lµ x ta cã:[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Al – 3e [/FONT]®[FONT=.VnTime] Al3+[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; Mg – 2e [/FONT]®[FONT=.VnTime] Mg2+&nbsp;&nbsp; tæng sè mol e nh&amp;shy;êng = 0,04. 3 + 0,06. 2 = 0,24[/FONT][FONT=.VnTime]&nbsp; &nbsp;&nbsp; S6+ + (6 – xe) [/FONT]®[FONT=.VnTime] S x&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; tæng sè mol e thu = (6 - x). 0,03 [/FONT][FONT=.VnTime]Theo quy t¾c b¶o toµn sè mol e: (6 - x). 0,03 = 0,24 [/FONT]®[FONT=.VnTime]&nbsp; x = – 2 [/FONT]®[FONT=.VnTime] s¶n phÈm lµ H2S[/FONT][FONT=.VnTime]b)&nbsp; N5+ + (5 – ye) [/FONT]®[FONT=.VnTime] S y&nbsp;&nbsp; t&amp;shy;¬ng tù: (5 - x). 0,03 = 0,24 [/FONT]®[FONT=.VnTime]&nbsp; x = – 3 [/FONT]®[FONT=.VnTime] s¶n phÈm lµ NH3[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; 4Mg +10 HNO3 [/FONT]®[FONT=.VnTime] 4Mg(NO3)2 + NH4NO&amp;shy;3 + 3H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; 8Al&nbsp;&nbsp; +30 HNO3 [/FONT]®[FONT=.VnTime] 8Al(NO3)3 + 3NH4NO&amp;shy;3 + 9H2O[/FONT][FONT=.VnTime]Theo pt: sè mol HNO3 ph¶n øng = 0,06. 2,5 + 0,04. 3,75 = 0,3 [/FONT]®[FONT=.VnTime] V[/FONT][FONT=.VnTime]HNO3 = 150 ml.[/FONT][FONT=.VnTime]Bµi 8.&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; Mét miÕng Mg bÞ oxihãa mét phÇn ®&amp;shy;îc chia lµm 2 phÇn b»ngnhau: [/FONT][FONT=.VnTime]- PhÇn 1 cho hßa tan hÕt trong dung dÞch HCl th× tho¸t ra 3,136 lÝt khÝ. C« c¹n dung dÞch thu ®&amp;shy;îc 14,25 gam chÊt r¾n A.[/FONT][FONT=.VnTime]- PhÇn 2 cho hßa tan hÕt trong dung dÞch HNO3 th× tho¸t ra 0,448 lÝt khÝ X nguyªn chÊt. C« c¹n dung dÞch thu ®&amp;shy;îc 23 gam chÊt r¾n B.[/FONT][FONT=.VnTime]a) TÝnh % sè mol Mg ®· bÞ oxihãa.(c¸c thÓ tÝch khÝ ®Òu ®o ë ®ktc)[/FONT][FONT=.VnTime]b) X¸c ®Þnh khÝ X. [/FONT][FONT=.VnArialH]Gi¶i: [/FONT][FONT=.VnTime]a)[/FONT][FONT=.VnArialH]&nbsp; [/FONT][FONT=.VnTime]sè[/FONT][FONT=.VnArialH] [/FONT][FONT=.VnTime]mol[/FONT][FONT=.VnArialH] [/FONT][FONT=.VnTime]khÝ H2 = 0,14 vµ sè mol khÝ X = 0,02[/FONT][FONT=.VnArialH]&nbsp; [/FONT][FONT=.VnArialH]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; [/FONT][FONT=.VnTime]2Mg + O[/FONT][FONT=.VnTime]2 [/FONT]®[FONT=.VnTime] 2MgO[/FONT][FONT=.VnArialH] [/FONT][FONT=.VnArialH]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; [/FONT][FONT=.VnTime]Mg + 2HCl [/FONT]®[FONT=.VnTime] MgCl2 + H[/FONT][FONT=.VnTime]2[/FONT]&amp;shy;[FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; MgO + 2HCl [/FONT]®[FONT=.VnTime] MgCl2 + H2O[/FONT][FONT=.VnTime]Theo pt: Mg ch&amp;shy;a bÞ oxihãa = 0,14 mol [/FONT]®[FONT=.VnTime] MgO =&nbsp; – 0,14 = 0,01 mol[/FONT][FONT=.VnTime]&nbsp; &nbsp;&nbsp; % Mg bÞ oxihãa =&nbsp; . 100% = 6,67%[/FONT][FONT=.VnTime]b)&nbsp; &nbsp; &nbsp; &nbsp; Mg + HNO3 [/FONT]®[FONT=.VnTime] Mg(NO3)2 + X + H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp;&nbsp; MgO + 2HNO3 [/FONT]®[FONT=.VnTime] Mg(NO3)2 + 2H2O[/FONT][FONT=.VnTime]BiÖn luËn:&nbsp; &nbsp; Theo tÝnh to¸n trªn sè mol Mg(NO3)2 = 0,15 [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; nªn l&amp;shy;îng Mg(NO3)2 = 0,15. 148 = 22,2 gam&nbsp; &lt; 23[/FONT][FONT=.VnTime]VËy trong B kh«ng chØ cã Mg(NO3)2 mµ cßn ph¶i cã NH4NO3 = = 0,01 mol[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 4Mg + 10HNO3 [/FONT]®[FONT=.VnTime] 4Mg(NO3)2 + NH4NO3 + 3H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 0,04&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 0,01[/FONT][FONT=.VnTime]&nbsp; &nbsp;&nbsp; (5x-2y)Mg + (12x-4y)HNO3 [/FONT]®[FONT=.VnTime] (5x-2y)Mg(NO3)2 + 2N[/FONT][FONT=.VnTime]xOy [/FONT]&amp;shy;[FONT=.VnTime] + (6x-2y)H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 0,14 – 0,04 = 0,1&nbsp; &nbsp; &nbsp;&nbsp; 0,02[/FONT][FONT=.VnTime]&nbsp;&nbsp; Theo pt:&nbsp; &nbsp;&nbsp; =&nbsp;&nbsp; [/FONT]®[FONT=.VnTime]&nbsp; 5x – 2y = 10 [/FONT]®[FONT=.VnTime]&nbsp; x = 2; y = 0 [/FONT]®[FONT=.VnTime] khÝ X lµ N2.[/FONT][FONT=.VnTime]&nbsp; Bµi 9:[/FONT][FONT=.VnTime]&nbsp; &nbsp; Cho 23,52g hçn hîp 3 kim lo¹i Mg, Fe, Cu vµo 200ml dung dÞch HNO3 3,4M khuÊy ®Òu thÊy tho¸t ra mét khÝ duy nhÊt h¬i nÆng h¬n kh«ng khÝ, trong dung dÞch cßn d&amp;shy; mét kim lo¹i ch&amp;shy;a tan hÕt, ®æ tiÐp tõ tõ dung dÞch H2SO4 5M vµo, chÊt khÝ trªn l¹i tho¸t ra cho dÕn khi kim lo¹i võa tan hÕt th× mÊt ®óng 44ml, thu ®&amp;shy;îc dd A. LÊy 1/2 dd A, cho dd NaOH cho ®Õn d&amp;shy; vµo, läc kÕt tña, röa råi nung ngoµi kh«ng khÝ ®Õn khèi l&amp;shy;îng kh«ng ®æi thu ®&amp;shy;îc chÊt r¾n B nÆng 15,6g.[/FONT][FONT=.VnTime] 1-TÝnh % sè mol mçi kim lo¹i trong hçn hîp.[/FONT][FONT=.VnTime] 2-TÝnh nång ®é c¸c ion (trõ ion H+‑, OH-) trong dung dÞch A.[/FONT][FONT=.VnArialH]Gi¶i: [/FONT][FONT=.VnTime]Gäi x, y, z lµ sè mol Mg, Fe, Cu trong hçn hîp, ta cã : [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 24x + 56y + 64z = 23,52&nbsp; &nbsp; &nbsp;&nbsp; [/FONT]«[FONT=.VnTime]&nbsp;&nbsp; 3x + 7y + 8z = 2,94&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (a) [/FONT][FONT=.VnTime]§ång cßn d&amp;shy; cã c¸c ph¶n øng:[/FONT][FONT=.VnTime]Cho e:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; NhËn e:[/FONT][FONT=.VnTime]Mg - 2e = Mg2+&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (1)&nbsp; &nbsp; &nbsp; &nbsp; NO3- + 3e + 4H+ = NO + 2H2O&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; (4)[/FONT][FONT=.VnTime] Fe - 3e = Fe3+&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; (2)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Cu + Fe3+ = Cu2+ + Fe2+&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; (5)[/FONT][FONT=.VnTime]Cu - 2e = Cu2+&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; (3)[/FONT][FONT=.VnTime]Ph&amp;shy;¬ng tr×nh ph¶n øng hoµ tan Cu d&amp;shy;:[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 3Cu + 4H2SO4 + 2NO3- = 3CuSO4 + SO42- + 2NO + H2O&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (6)[/FONT][FONT=.VnTime]Tõ Pt (6) tÝnh ®&amp;shy;îc sè mol Cu d&amp;shy;: =&nbsp; =&nbsp; 0,165 mol[/FONT][FONT=.VnTime]Theo c¸c ph&amp;shy;¬ng tr×nh (1), (2), (3), (4), (5): sè mol cho b»ng sè mol e nhËn:[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2(x + y + z – 0,165) = [/FONT][[FONT=.VnTime]3,4.0,2 – 2(x + y + z – 0,165)[/FONT]][FONT=.VnTime].3[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; [/FONT]®[FONT=.VnTime] x + y + z = 0,255 + 0,165 = 0,42&nbsp;&nbsp; (b)[/FONT][FONT=.VnTime]Tõ khèi l&amp;shy;îng c¸c oxit MgO; Fe2O3; CuO, cã ph&amp;shy;¬ng tr×nh: .40 + .160 + . 80 = 15,6&nbsp;&nbsp; (c)[/FONT][FONT=.VnTime][/FONT][FONT=.VnTime]HÖ ph&amp;shy;¬ng tr×nh rót ra tõ (a), (b), (c):&nbsp; &nbsp; &nbsp;&nbsp; 3x + 7y + 8z = 2,94[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; x + y + z = 0,42[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; x + 2y + 2z = 0,78[/FONT][FONT=.VnTime]Gi¶i ®&amp;shy;îc:&nbsp; x = 0,06;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; y = 0,12;&nbsp; &nbsp; &nbsp; &nbsp; z = 0,24.[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; [/FONT]%[FONT=.VnTime] l&amp;shy;îng Mg =&nbsp; 6,12[/FONT]%[FONT=.VnTime]&nbsp; ;&nbsp; [/FONT]%[FONT=.VnTime] l&amp;shy;îng Fe =&nbsp; 28,57[/FONT]%[FONT=.VnTime]&nbsp; ;&nbsp; [/FONT]%[FONT=.VnTime] l&amp;shy;îng Cu = 65,31[/FONT]%[FONT=.VnTime]2/ TÝnh nång ®é c¸c ion trong dd A (trõ H+, OH-)&nbsp; &nbsp;&nbsp; [/FONT][[FONT=.VnTime]Mg2+[/FONT]][FONT=.VnTime] =&nbsp; = 0,246 M[/FONT][FONT=.VnTime]&nbsp; &nbsp;&nbsp; [/FONT][[FONT=.VnTime]Cu2+[/FONT]][FONT=.VnTime] =&nbsp; 0,984 M&nbsp; ;&nbsp; &nbsp;&nbsp; [/FONT][[FONT=.VnTime]Fe2+[/FONT]][FONT=.VnTime] =&nbsp; 0,492 M ;&nbsp; &nbsp; [/FONT][[FONT=.VnTime]SO[/FONT][FONT=.VnTime]42-[/FONT]][FONT=.VnTime] =&nbsp; 0,9 M ;&nbsp; &nbsp; [/FONT][[FONT=.VnTime]NO[/FONT][FONT=.VnTime]3-[/FONT]][FONT=.VnTime] =&nbsp; 1,64 M[/FONT][FONT=.VnTime]&nbsp; Bµi 10:[/FONT][FONT=.VnTime]&nbsp; &nbsp;&nbsp; Hoµ tan hoµn toµn hçn hîp FeS vµ FeCO3 b»ng dung dÞch HNO3 ®Æc, nãng thu ®&amp;shy;îc hçn hîp (B) gåm 2 khÝ X vµ Y cã tû khèi ®èi víi H2 lµ 22,8 vµ cßn dung dÞch (A) cã pH &lt; 3.[/FONT][FONT=.VnTime]a)TÝnh tû lÖ sè mol c¸c muèi Fe2+ trong hçn hîp ban ®Çu.[/FONT][FONT=.VnTime]b)Lµm l¹nh hçn hîp khÝ (B) xuèng nhiÖt ®é thÊp h¬n ®&amp;shy;îc hçn hîp (B[/FONT]¢[FONT=.VnTime]) gåm 3 khÝ X,Y,Z cã tû khèi so víi H2 b»ng 28,5. TÝnh phÇn tr¨m theo thÓ tÝch cña hçn hîp khÝ (B[/FONT]¢[FONT=.VnTime]).[/FONT][FONT=.VnTime]c) [/FONT][FONT=.VnTimeH]ë[/FONT][FONT=.VnTime] -11oC hçn hîp (B[/FONT]¢[FONT=.VnTime]) chuyÓn sang (B[/FONT]²[FONT=.VnTime]) gåm 2 khÝ. TÝnh tû khèi cña (B[/FONT]²[FONT=.VnTime]) so víi H2.[/FONT][FONT=.VnArialH]Gi¶i:[/FONT][FONT=.VnTime] a) PT pø:&nbsp; &nbsp; &nbsp;&nbsp; FeS + 12HNO[/FONT][FONT=.VnTime]3 [/FONT]®[FONT=.VnTime] Fe(NO3)3 + H2SO4 + 9NO[/FONT][FONT=.VnTime]2[/FONT]&amp;shy;[FONT=.VnTime] + 5H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; FeCO3 + 4HNO3 [/FONT]®[FONT=.VnTime] Fe(NO3)3 + CO2 + NO[/FONT][FONT=.VnTime]2[/FONT]&amp;shy;[FONT=.VnTime] + 2H2O[/FONT][FONT=.VnTime]§Æt:&nbsp; nFeS = a mol ,&nbsp; nFeCO3 = b mol&nbsp;&nbsp; [/FONT]®[FONT=.VnTime]&nbsp; nNO2 = 9a + b&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; vµ nCO2 = b&nbsp;&nbsp; [/FONT][FONT=.VnTime]Ta cã :&nbsp; &nbsp; = 22,8&nbsp; [/FONT]®[FONT=.VnTime]&nbsp; a:b = 1:3 [/FONT]®[FONT=.VnTime]&nbsp; n FeS : n FeCO3 = 1: 3&nbsp;&nbsp; (b = 3a)[/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime]b) Lµm l¹nh B cã ph¶n øng : 2NO2&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; N2O4&nbsp; khi ®ã M (N2O4) = 92 lµm t¨ng vµ = 57[/FONT][FONT=.VnTime]Gäi x lµ sè mol N2O4 trong hçn hîp B, vËy B[/FONT]¢[FONT=.VnTime] gåm:[/FONT][FONT=.VnTime]NO2 = (9a + b) – 2x = 4b - 2x&nbsp; ;&nbsp;&nbsp; N2O4 = x vµ CO2 = b [/FONT]®[FONT=.VnTime]&nbsp; = 57[/FONT]®[FONT=.VnTime] b = x&nbsp; [/FONT]®[FONT=.VnTime] Tæng B’ = 4b gåm NO2 = 2b [/FONT]~[FONT=.VnTime] 50[/FONT]%[FONT=.VnTime] ; N2O4 = b [/FONT]~[FONT=.VnTime]&nbsp; 25[/FONT]%[FONT=.VnTime] ;&nbsp; CO2 = b [/FONT]~[FONT=.VnTime]&nbsp; 25[/FONT]%[FONT=.VnTime]c) ë – 110C ph¶n øng dime ho¸ x¶y ra hoµn toµn, B[/FONT]¢¢[FONT=.VnTime] gåm N2O4 = 2b&nbsp; vµ CO2 = b[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; tØ khèi so víi hidro =&nbsp; = 38[/FONT][FONT=.VnTime] [/FONT][FONT=.VnTime]Bµi 11:[/FONT][FONT=.VnTime]&nbsp; &nbsp; Cho tõ tõ khÝ CO qua èng chøa 6,4gam CuO ®un nãng. KhÝ ra khái èng ®&amp;shy;îc hÊp thô hoµn toµn b»ng 150ml dung dÞch n&amp;shy;íc v«i trong nång ®é 0,1M thÊy t¸ch ra 1,0 gam kÕt tña tr¾ng, ®un s«i phÇn n&amp;shy;íc läc l¹i thÊy cã vÈn ®ôc. ChÊt r¾n cßn l¹i trong èng ®&amp;shy;îc cho vµo 500ml dung dÞch HNO3 0,32M tho¸t ra V1 lÝt khÝ NO2 nÕu thªm 760ml dung dÞch HCl 1,333M vµo dung dÞch sau ph¶n øng th× l¹i tho¸t ra thªm V2 lÝt khÝ NO n÷a. NÕu tiÕp tôc thªm 24 gam Mg th× thÊy tho¸t ra V3 lÝt khÝ hçn hîp khÝ N2 vµ H2, läc dung dÞch cuèi cïng thu ®&amp;shy;îc chÊt r¾n X.[/FONT][FONT=.VnTime]a/ ViÕt ph&amp;shy;¬ng tr×nh ph¶n øng vµ tÝnh V1&amp;shy;,V2,V3(®ktc).[/FONT][FONT=.VnTime]b/ TÝnh thµnh phÇn X( gi¶ thiÕt c¸c ph¶n øng x¶y ra hoµn toµn).[/FONT][FONT=.VnArialH]Gi¶i:[/FONT][FONT=.VnTime]a)[/FONT][FONT=.VnArialH]&nbsp; &nbsp; &nbsp; [/FONT][FONT=.VnTime] CO + CuO [/FONT]®[FONT=.VnTime] Cu + CO2&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; sè mol CO2 = 0,02[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; CO2 + Ca(OH)2 [/FONT]®[FONT=.VnTime]&nbsp; CaCO3 + H2O&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; sè mol Cu = 0,02[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 2CO2 + Ca(OH)2 [/FONT]®[FONT=.VnTime]&nbsp; Ca(HCO3)2&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; sè mol CuO = 0,06[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; Ca(HCO3)2 [/FONT]®[FONT=.VnTime] CaCO3 + CO[/FONT][FONT=.VnTime]2[/FONT]&amp;shy;[FONT=.VnTime]+ H2O&nbsp; &nbsp;&nbsp; sè mol HNO3 = 0,16[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; CuO + 2H+ [/FONT]®[FONT=.VnTime] Cu2+ + H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; 3Cu + 8H+ + 2NO3– [/FONT]®[FONT=.VnTime] 3Cu2+ + 2NO[/FONT]&amp;shy;[FONT=.VnTime] + 4H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; 0,015&nbsp; &nbsp; 0,04&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 0,01[/FONT][FONT=.VnTime]Theo pt:&nbsp; &nbsp;&nbsp; V1 = 0,01. 22,4 = 0,224 lÝt ;&nbsp; &nbsp; &nbsp;&nbsp; V2 =&nbsp; V1 = 0,07467 lÝt[/FONT][FONT=.VnTime]Khi thªm mol HCl, ph¶n øng l¹i tiÕp tôc x¶y ra[/FONT][FONT=.VnTime]Thªm 1 mol Mg:&nbsp; &nbsp; &nbsp; 5Mg + 12H+ + 2NO3– [/FONT]®[FONT=.VnTime] 5Mg2+ + N[/FONT][FONT=.VnTime]2[/FONT]&amp;shy;[FONT=.VnTime] + 6H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; Mg + Cu2+ [/FONT]®[FONT=.VnTime]&nbsp;&nbsp; Mg2+ + Cu[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Mg + 2H+ [/FONT]®[FONT=.VnTime]&nbsp;&nbsp; Mg2+ + H[/FONT][FONT=.VnTime]2[/FONT]&amp;shy;[FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; V3 =&nbsp; V[/FONT][FONT=.VnTime]N2 + VH2 [/FONT][FONT=.VnTime]mµ sè mol N2 = (0,16 - ) =&nbsp; &nbsp; vµ&nbsp; &nbsp; H2 = [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp;&nbsp; nªn V3 = 22,4 = 2,9867 lÝt[/FONT][FONT=.VnTime]b)&nbsp; Suy ra&nbsp;&nbsp; Cu = 0,08. 64 = 5,12 gam chiÕm 30,19% [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp;&nbsp; vµ Mg = .24 = 11,84 gam chiÕm 69,81%[/FONT][FONT=.VnTime]Bµi 12:[/FONT][FONT=.VnTime]1/Hoµ tan khÝ SO2 vµo H2O cã c¸c c©n b»ng sau:[/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; SO2&nbsp; +&nbsp; H2O&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; H2SO3&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; (1)[/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; H2SO3&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; H+&nbsp; +&nbsp; HSO3–&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (2)[/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; HSO3&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; H+&nbsp; +&nbsp; SO32–&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (3)[/FONT][FONT=.VnTime]Nång ®é SO2 thay ®æi nh&amp;shy; thÕ nµo? (gi¶i thÝch) khi lÇn l&amp;shy;ît t¸c ®éng nh÷ng yÕu tè sau:[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; a)§un nãng dung dÞch.[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; b)Thªm dung dÞch HCl.[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; c)Thªm dung dÞch NaOH.[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; d)Thªm dung dÞch KMnO4.[/FONT][FONT=.VnArialH]Gi¶i: [/FONT][FONT=.VnTime] a)§un nãng dd: SO2 tho¸t khái dd [/FONT]®[FONT=.VnTime] [/FONT][[FONT=.VnTime] SO[/FONT][FONT=.VnTime]2 [/FONT]][FONT=.VnTime] trong dung dÞch gi¶m.[/FONT][FONT=.VnTime]&nbsp;&nbsp; b)Thªm dd HCl:&nbsp; Cb»ng&nbsp; (2) (3) chuyÓn sang tr¸i [/FONT]®[FONT=.VnTime]C©n b»ng (1) chuyÓn sang tr¸i[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; [/FONT]®[FONT=.VnTime] [/FONT][[FONT=.VnTime] H2SO[/FONT][FONT=.VnTime]3 [/FONT]][FONT=.VnTime] t¨ng [/FONT]®[FONT=.VnTime] [/FONT][[FONT=.VnTime] SO[/FONT][FONT=.VnTime]2 [/FONT]][FONT=.VnTime] t¨ng.[/FONT][FONT=.VnTime]&nbsp; &nbsp; c)Thªm NaOH:&nbsp; &nbsp;&nbsp; NaOH&nbsp; +&nbsp; SO2&nbsp; =&nbsp; NaHSO3&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 2 NaOH&nbsp; +&nbsp; SO2&nbsp; =&nbsp; Na2SO3&nbsp; +&nbsp; H2O&nbsp;&nbsp; [/FONT][FONT=.VnTime]Cb»ng&nbsp; (2) (3) chuyÓn sang ph¶i [/FONT]®[FONT=.VnTime]C©n b»ng (1) chuyÓn sang ph¶i&nbsp;&nbsp; [/FONT]®[FONT=.VnTime]&nbsp; [/FONT][[FONT=.VnTime]SO[/FONT][FONT=.VnTime]2 [/FONT]][FONT=.VnTime]&nbsp; gi¶m[/FONT][FONT=.VnTime]&nbsp; &nbsp; d)Thªm KMnO4.&nbsp; Cã pø:&nbsp; &nbsp; [/FONT]®[FONT=.VnTime]&nbsp;&nbsp; lµm&nbsp; gi¶m [/FONT][[FONT=.VnTime] SO[/FONT][FONT=.VnTime]2 [/FONT]][FONT=.VnTime] .[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 5SO2&nbsp; +&nbsp; 2KMnO4&nbsp; +&nbsp; 2H2O&nbsp; =&nbsp; 2H2SO4&nbsp; +&nbsp; 2MnSO4&nbsp; +&nbsp; K2SO4[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; SO32– +&nbsp; 2KMnO4&nbsp; +&nbsp; 2H2O&nbsp; =&nbsp; H2SO4&nbsp; + K2MnO4&nbsp; +&nbsp; MnO42–&nbsp; &nbsp; &nbsp; &nbsp; [/FONT][FONT=.VnTime]Bµi 13:[/FONT][FONT=.VnTime]1- DÉn ra 5 ph&amp;shy;¬ng tr×nh ph¶n øng kh¸c nhau cã t¹o ra khÝ NO2.[/FONT][FONT=.VnTime]2-a)TÝnh pH cña dung dÞch HCl nång ®é 0,5.10-7 mol/lÝt.[/FONT][FONT=.VnTime]&nbsp; b)TÝnh pH cña dung dÞch X ®&amp;shy;îc t¹o thµnh khi trén 200ml dung dÞch HA 0,1M (Ka = 10-3.75) víi 200ml dung dÞch KOH 0.05M; pH cña dung dÞch X thay ®æi nh&amp;shy; thÕ nµo khi thªm 10-3 mol HCl vµo dung dÞch X.[/FONT][FONT=.VnArialH]Gi¶i:[/FONT][FONT=.VnTime]&nbsp; &nbsp;&nbsp; 1/&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 2HNO3&nbsp; =2NO2&nbsp; +&nbsp; 1/2O2&nbsp; + 2H2O&nbsp; &nbsp;&nbsp; [/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; Cu + 4HNO&amp;shy;3 = Cu(NO3)2 + 2NO2[/FONT]&amp;shy;[FONT=.VnTime] + 2H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2AgNO3 = 2Ag + O2[/FONT]&amp;shy;[FONT=.VnTime] + 2NO[/FONT][FONT=.VnTime]2[/FONT]&amp;shy;[FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; C + 4HNO3 = CO2 + 4NO2[/FONT]&amp;shy;[FONT=.VnTime] + 2H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 2NO + O2 = 2NO2[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 2/a) [/FONT][[FONT=.VnTime] H[/FONT][FONT=.VnTime]+[/FONT]][FONT=.VnTime] . 0,5.10-7 do nång ®é nhá [/FONT]®[FONT=.VnTime] ph¶i tÝnh ®Õn c©n b»ng cña H2O&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; [/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; H2O&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; H+ + OH –[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; HCl&nbsp; &nbsp; &nbsp; &nbsp; [/FONT]®[FONT=.VnTime]&nbsp; &nbsp;&nbsp; H+ + Cl –[/FONT][FONT=.VnTime] Theo ®Þnh luËt b¶o toµn ®iÖn tÝch:&nbsp; [/FONT][[FONT=.VnTime] H[/FONT][FONT=.VnTime]+[/FONT]][FONT=.VnTime] = [/FONT][[FONT=.VnTime] Cl[/FONT][FONT=.VnTime]-[/FONT]][FONT=.VnTime] + [/FONT][[FONT=.VnTime]OH[/FONT][FONT=.VnTime]-[/FONT]][FONT=.VnTime]&nbsp; [/FONT]®[FONT=.VnTime]&nbsp;&nbsp; [/FONT][[FONT=.VnTime] H+[/FONT]][FONT=.VnTime] = 0,5.10-7 + [/FONT][FONT=.VnTime]&nbsp; &nbsp;&nbsp; [/FONT]®[FONT=.VnTime] [/FONT][[FONT=.VnTime] H+[/FONT]][FONT=.VnTime]2 –&nbsp; 0,5.10 [/FONT][FONT=.VnTime]–7[/FONT][[FONT=.VnTime] H[/FONT][FONT=.VnTime]+[/FONT]][FONT=.VnTime] – 10-14 = 0.&nbsp; Gi¶i ®&amp;shy;îc:&nbsp; [/FONT][[FONT=.VnTime] H[/FONT][FONT=.VnTime]+[/FONT]][FONT=.VnTime] = 1,28.10[/FONT][FONT=.VnTime]-7 [/FONT]®[FONT=.VnTime] pH [/FONT]»[FONT=.VnTime] 6,9[/FONT][FONT=.VnTime]b)&nbsp;&nbsp; nHA = 0,1.0,2 = 0,02 mol&nbsp;&nbsp; ;&nbsp; &nbsp; &nbsp;&nbsp; nKOH = 0,05.0,2 = 0,01 mol[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; KOH + HA [/FONT]®[FONT=.VnTime] KA + H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; 0,01 [/FONT]®[FONT=.VnTime] 0,01[/FONT]®[FONT=.VnTime] 0,01&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Theo ph&amp;shy;¬ng tr×nh HA cßn d&amp;shy; = 0,01 mol[/FONT][FONT=.VnTime]Trong d2 X: CHA = CKA =&nbsp; = 0,025M.&nbsp; XÐt c¸c c©n b»ng sau: [/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime]&nbsp; H2O&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; H+ + OH-&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; KW&nbsp; = 10-14&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; (1) [/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime]&nbsp;&nbsp; HA&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; H+ + A-&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; KHA = 10-375&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (2)[/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime]&nbsp;&nbsp; A- + H2O&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; HA + OH-&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; KB = KHA-1.&nbsp; KW = 10-10,25&nbsp; &nbsp; (3) [/FONT][FONT=.VnTime]So s¸nh (1) víi (2)&nbsp; [/FONT]®[FONT=.VnTime]&nbsp; KHA &gt;&gt; K[/FONT][FONT=.VnTime]W [/FONT]®[FONT=.VnTime] bá qua (1)&nbsp; &nbsp; &nbsp;&nbsp; [/FONT][FONT=.VnTime]So s¸nh (2) víi (3)&nbsp; [/FONT]®[FONT=.VnTime]&nbsp; KHA &gt;&gt; K[/FONT][FONT=.VnTime]B [/FONT]®[FONT=.VnTime] bá qua(3) [/FONT]®[FONT=.VnTime] Dung dÞch X lµ dung dÞch ®Öm axit[/FONT][FONT=.VnTime]cã pH = pKa + lg&nbsp; = 3,75 + lg&nbsp; = 3,75[/FONT][FONT=.VnTime] [/FONT]*[FONT=.VnTime] Khi thªm 10-3 mol HCl:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; KA&nbsp;&nbsp; +&nbsp;&nbsp; HCl&nbsp;&nbsp; [/FONT]®[FONT=.VnTime]&nbsp;&nbsp; KCl + HA[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 0,001 [/FONT]¬[FONT=.VnTime] 0,001 [/FONT]®[FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 0,001&nbsp; &nbsp;&nbsp; (mol)[/FONT][FONT=.VnTime]&nbsp; [/FONT][[FONT=.VnTime]HA[/FONT]][FONT=.VnTime] =&nbsp; = 0,0275 M&nbsp; vµ&nbsp;&nbsp; [/FONT][[FONT=.VnTime]KA[/FONT]][FONT=.VnTime] =&nbsp; = 0,0225M .[/FONT][FONT=.VnTime]Dung dÞch thu ®&amp;shy;îc vÉn lµ dung dÞch ®Öm axit nªn pH = 3,75 + lg&nbsp; = 3,66[/FONT][FONT=.VnTime]Bµi 14: [/FONT][FONT=.VnTime]&nbsp; &nbsp; 1/ Dung dÞch A gåm c¸c cation: NH4+ ; Na+ ; Ba2+ vµ 1 anion X cã thÓ lµ mét trong c¸c&nbsp; anion sau: CH3COO – ; NO3–; SO42– ; CO32– ; PO43– . Hái X lµ anion nµo? BiÕt r»ng dung dÞch A cã pH = 5 .[/FONT][FONT=.VnTime]&nbsp; &nbsp; 2/ Thªm NaOH d&amp;shy; vµo dung dÞch CuSO4 ,thªm tiÕp NH4NO3 vµo dung dÞch ®Õn d&amp;shy; cã hiÖn t&amp;shy;îng g× x¶y ra? ViÕt ph&amp;shy;¬ng tr×nh ph¶n øng.[/FONT][FONT=.VnTime]&nbsp; &nbsp; 3/ Cho c¸c dung dÞch sau: Al(NO3)3 ; ZnCl2 ; Na2CO3 ; MgSO4 ; NH4NO3 . Kh«ng dïng thªm thuèc thö h·y nªu c¸ch nhËn ra tõng dung dÞch, viÕt ph&amp;shy;¬ng tr×nh ph¶n øng d¹ng ion.[/FONT][FONT=.VnArialH]Gi¶i:&nbsp; &nbsp; [/FONT][FONT=.VnTime]1/ X lµ NO[/FONT][FONT=.VnTime]3–&nbsp; v× NH4NO3: m«i tr&amp;shy;êng axit pH&lt; 7.&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; [/FONT][FONT=.VnTime]2/ + cã kÕt tña mµu xanh: Cu2+ + 2OH- = Cu(OH)2[/FONT]¯[FONT=.VnTime]&nbsp; &nbsp; + cã khÝ mïi khai : NH4+ + OH- = NH3[/FONT]&amp;shy;[FONT=.VnTime] + H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; + kÕt tña tan t¹o dung dÞch xanh thÉm:&nbsp; &nbsp; Cu(OH)2 + 4NH3 = [Cu(NH3)4]2+ + 2OH-[/FONT][FONT=.VnTime]3/ LÊy mçi dung dÞch mét Ýt cho lÉn l&amp;shy;ît vµo c¸c dung dÞch cßn l¹i hiÖn t&amp;shy;îng ®&amp;shy;îc ghi trong b¶ng sau:[/FONT][FONT=.VnTime] [/FONT][FONT=.VnTime]Al(NO3)3[/FONT][FONT=.VnTime]ZnCl2[/FONT][FONT=.VnTime]Na2CO3[/FONT][FONT=.VnTime]MgSO4[/FONT][FONT=.VnTime]NH4NO3[/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime][/FONT][FONT=.VnTime]Al(NO3)3[/FONT][FONT=.VnTime] [/FONT][FONT=.VnTime]-[/FONT]¯[FONT=.VnTime], [/FONT]¯[FONT=.VnTime]tan (1)[/FONT][FONT=.VnTime]-[/FONT][FONT=.VnTime]-[/FONT][FONT=.VnTime]ZnCl2[/FONT][FONT=.VnTime]-[/FONT][FONT=.VnTime] [/FONT]¯[FONT=.VnTime], [/FONT]¯[FONT=.VnTime]tan (2)[/FONT][FONT=.VnTime]-[/FONT][FONT=.VnTime]-[/FONT][FONT=.VnTime]Na2CO3[/FONT]¯[FONT=.VnTime], [/FONT]¯[FONT=.VnTime]tan[/FONT]¯[FONT=.VnTime], [/FONT]¯[FONT=.VnTime]tan[/FONT][FONT=.VnTime] [/FONT]¯[FONT=.VnTime]tr¾ng[/FONT]&amp;shy;[FONT=.VnTime](khai)[/FONT][FONT=.VnTime]MgSO4[/FONT][FONT=.VnTime]-[/FONT][FONT=.VnTime]-[/FONT]¯[FONT=.VnTime], [/FONT]¯[FONT=.VnTime]tan (3)[/FONT][FONT=.VnTime] [/FONT][FONT=.VnTime]-[/FONT][FONT=.VnTime]NH4NO3[/FONT][FONT=.VnTime]-[/FONT][FONT=.VnTime]-[/FONT]&amp;shy;[FONT=.VnTime] (khai)(4)[/FONT][FONT=.VnTime]-[/FONT][FONT=.VnTime] [/FONT][FONT=.VnTime]KÕt luËn[/FONT]¯[FONT=.VnTime], [/FONT]¯[FONT=.VnTime]tan[/FONT]¯[FONT=.VnTime], [/FONT]¯[FONT=.VnTime]tan[/FONT][FONT=.VnTime]3[/FONT]¯[FONT=.VnTime],[/FONT]&amp;shy;¯&amp;shy;[FONT=.VnTime]-NÕu dung dÞch cho 3 tr&amp;shy;êng hîp kÕt tña vµ 1 tr&amp;shy;êng hîp cã khÝ mïi khai tho¸t ra lµ dung dÞch Na2CO3:[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; Al3+ + 3OH- = Al(OH)3[/FONT]¯[FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (CO32- + H2O [/FONT]Û[FONT=.VnTime] HCO3- + OH-)[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; Al(OH)3 + OH- = AlO2- + 2H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; Zn2+ + 2OH- = Zn(OH)2[/FONT][FONT=.VnTime]&nbsp; &nbsp; Zn(OH)2 + 2OH- =ZnO22- + 2H2O[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Mg2+ + 2OH- = Mg(OH)2&nbsp; &nbsp; &nbsp;&nbsp; (kh«ng viÕt Mg2+ +&nbsp; CO32- = MgCO3)[/FONT][FONT=.VnTime]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; NH4+ + OH-&nbsp; = NH3 + H2O&nbsp; &nbsp;&nbsp; [/FONT][FONT=.VnTime]- NÕu dung dÞch chØ cho 1 kÕt tña vµ kÕt tña kh«ng tan trong thuèc thö d&amp;shy; lµ dung dÞch MgSO4.[/FONT][FONT=.VnTime]- NÕu dung dÞch chØ cho 1 tr&amp;shy;êng hîp cã khÝ mïi khai tho¸t ra lµ dung dÞch NH4NO3.[/FONT][FONT=.VnTime]- 2 dung dÞch cßn l¹i cã hiÖn t&amp;shy;îng gièng nhau lµ Al(NO3)3 vµ ZnCl2[/FONT][FONT=.VnTime]&nbsp; Dïng dung dÞch&nbsp; NH4NO3 nhËn ra kÕt tña Zn(OH)2: kÕt tña tan trong dung dÞch NH4NO3&nbsp; [/FONT]®[FONT=.VnTime] dung dÞch t&amp;shy;¬ng øng lµ ZnCl2 cßn l¹i lµ dung dÞch Al(NO3)3[/FONT][FONT=.VnTime]&nbsp;&nbsp; NH4+&nbsp;&nbsp; [/FONT]Û[FONT=.VnTime]&nbsp; NH3 + H+&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; vµ&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Zn(OH)2 + 4NH3 = [Zn(NH3)4]2+ + 2OH-[/FONT]</blockquote>

[QUOTE="copemuadong_dn9, post: 72431, member: 74949"] [SIZE=4][FONT=.VnTime][B][U][COLOR=blue]Bµi 1.[/COLOR][/U][/B][U][COLOR=blue] [/COLOR][/U][/FONT][/SIZE] [COLOR=blue][SIZE=4][FONT=.VnTime] Hçn hîp X gåm 2 kim lo¹i Al vµ Cu. Cho 18,2 gam X vµo 100ml dung dÞch Y chøa H2SO4 12M vµ HNO3 2M, ®un nãng t¹o ra dung dÞch Z vµ 8,96 lÝt (®ktc) hçn hîp T gåm NO vµ khÝ D kh«ng mµu. Hçn hîp T cã tû khèi so víi hidro = 23,5. [/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime]TÝnh khèi lîng mçi kim lo¹i trong hçn hîp ®Çu vµ lîng mçi muèi trong dung dÞch Z[/FONT][/SIZE][/COLOR] [SIZE=4][FONT=.VnTime] [/FONT][U][FONT=.VnArialH]Gi¶i:[/FONT][/U][FONT=.VnArialH] [/FONT][FONT=.VnTime] M[/FONT][/SIZE][FONT=.VnTime]T[SIZE=4] =23,5 . 2 = 47 [/SIZE][/FONT][SIZE=4][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] M[/FONT][/SIZE][FONT=.VnTime]NO[SIZE=4] = 30 < 47 < M[/SIZE]D[SIZE=4]. [/SIZE][/FONT][SIZE=4][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] D lµ SO[/FONT][/SIZE][FONT=.VnTime]2[SIZE=4] = 64[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] Suy ra sè mol NO = 0,2 mol vµ SO2 = 0,2 mol[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]Thùc chÊt ph¶n øng theo c¸c ph¬ng tr×nh sau:[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] Al – 3e [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Al3+. Víi sè mol Al = x vµ sè mol Cu = y[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Cu – 2e [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Cu2+. Tæng sè e nhêng = 3x + 2y[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] NO3- + 4H+ + 3e [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] NO + 2H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] SO42– + 4H+ + 2e [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] SO2 + 2H2O (Tæng sè mol e thu = 0,6 + 0,4 = 1 )[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]Sè mol H+ trong Y = 1,2 . 2 + 0,2 = 2,6 > sè mol H+dù ph¶n øng = (0,2 + 0,2) 4= 1,6 [/SIZE][/FONT] [FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4]Nªn kim lo¹i tan hÕt.[/SIZE][/FONT] [FONT=.VnTime][SIZE=4]VËy ta cã hÖ ph¬ng tr×nh: 27x + 64y = 18,2[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] 3x + 2y = 1 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] gi¶i pt cho x = y = 0,2[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]V× NO3– ph¶n øng = NO3– trong Y nªn dung dÞch Z kh«ng cã NO3–‑ vµ chØ cã Al3+, Cu2+, SO42–. Lîng Al2(SO4)3 = . 342 = 34,2 gam[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] Lîng CuSO4 = 0,2 . 160 = 32 gam[/FONT][/SIZE] [SIZE=4][FONT=.VnTime][B][U][COLOR=blue]Bµi 2.[/COLOR][/U][/B][U][COLOR=blue] [/COLOR][/U][/FONT][/SIZE] [COLOR=blue][SIZE=4][FONT=.VnTime] Hßa tan a gam hçn hîp Cu vµ Fe (trong ®ã Fe chiÕm 30% vÒ khèi lîng) b»ng 50 ml dung dÞch HNO3 63% (d = 1,38 g/ml), khuÊy ®Òu cho tíi ph¶n øng hoµn toµn thu ®îc chÊt r¾n A c©n nÆng 0,75a gam , dung dÞch B vµ 6,104 lÝt hçn hîp khÝ NO vµ NO2 (®ktc). Hái c« c¹n dung dÞch B thu ®îc bao nhiªu gam muèi khan?[/FONT][/SIZE][/COLOR] [SIZE=4][U][FONT=.VnArialH]Gi¶i:[/FONT][/U][FONT=.VnArialH] [/FONT][FONT=.VnTime]PhÇn hçn hîp tan = 0,25a gam < 0,3a gam nªn Fe cßn d vµ Cu cha ph¶n øng[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] C¸c ph¶n øng: Fe + 4 HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Fe(NO3)3 + NO [/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime]+ 2H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Fe + 6 HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Fe(NO3)3 + 3NO[/FONT][FONT=.VnTime]2 [/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime]+ 3H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Do Fe d nªn cßn c¸c ph¶n øng: Fe + 2Fe(NO3)3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 3Fe(NO3)2 [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Tøc lµ trong dung dÞch B chØ cã Fe(NO3)2[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Vµ lîng muèi = lîng Fe + lîng NO3– [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Mµ sè mol NO3– cßn l¹i trong B = tæng sè mol NO3– – sè mol khÝ [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] = 0,69 – 0,2725 = 0,4175 mol[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] VËy lîng muèi khan = . 180 = 37,575 gam[/FONT][/SIZE] [SIZE=4][FONT=.VnTime][B][U][COLOR=blue]Bµi 3.[/COLOR][/U][/B][U][COLOR=blue] [/COLOR][/U][/FONT][/SIZE] [COLOR=blue][SIZE=4][FONT=.VnTime] Hßa tan 17,4 gam hçn hîp 3 kim lo¹i Al, Cu, Fe trong dung dÞch HCl d thÊy tho¸t ra 8,96 lÝt khÝ(®ktc). NÕu cho 34,8 gam hçn hîp trªn t¸c dông víi dung dÞch CuSO4 d råi läc chÊt r¾n t¹o ra hßa tan b»ng HNO3 th× tho¸t ra 26,88 lÝt khÝ (®ktc) cã tû khèi so víi oxi = 1,27. ViÕt c¸c ph¬ng tr×nh ph¶n øng vµ tÝnh thµnh phÇn hçn hîp ban ®Çu. [/FONT][/SIZE][/COLOR] [SIZE=4][FONT=.VnArialH]Gi¶i: [/FONT][FONT=.VnTime]2Al + 6 HCl [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 2AlCl3 + 3H[/FONT][FONT=.VnTime]2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] (1)[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Fe + 2 HCl [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] FeCl2 + H[/FONT][FONT=.VnTime]2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] (2)[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 2Al + 3CuSO4 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Al2(SO4)3 + 3Cu[/FONT][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime] (3) [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Fe + CuSO4 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] FeSO4 + Cu[/FONT][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime] (4)[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]KhÝ cã 30 < M[/SIZE]TB[SIZE=4] = 40,64 < 46 lµ hçn hîp NO vµ NO2 (cÇn chó ý lµ trong c¸c s¶n phÈm chøa N t¹o ra khi t¸c dông víi HNO3 ngoµi muèi, Cu chØ t¹o NO vµ NO2 )[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] Cu + 4HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Cu(NO3)2 + 2NO[/FONT][FONT=.VnTime]2 [/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime]+ 2H2O (5)[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 3Cu + 8HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 3Cu(NO3)2 + 2NO [/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime]+ 4H2O (6)[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]Tõ M[/SIZE]TB[SIZE=4] = 40,64 vµ tæng sè mol = 1,2 tÝnh ®îc NO = 0,4 mol vµ NO2= 0,8 mol[/SIZE][/FONT] [FONT=.VnTime][SIZE=4]Theo pt (5), (6) Cu = 0,4 + 0,6 = 1,0 mol[/SIZE][/FONT] [SIZE=4][FONT=.VnTime]So s¸nh (1), (2) víi (3), (4) thÊy sè mol Cu[/FONT][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime] = H2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime]= 0,8 mol (TN 2 lÊy lîng gÊp ®«i)[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]Suy ra Cu ban ®Çu = = 0,1 mol [/FONT][FONT=Symbol][FONT=Symbol]~[/FONT][/FONT][FONT=.VnTime] 6,4 gam [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Al + Fe = 11 gam[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]Tõ hÖ pt vÒ tæng lîng Al + Fe vµ tæng sè mol H2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] tÝnh ®îc Al = 5,4 g vµ Fe = 5,6 g[/FONT][/SIZE] [SIZE=4][FONT=.VnTime][B][U][COLOR=blue]Bµi 4.[/COLOR][/U][/B][U][COLOR=blue] [/COLOR][/U][/FONT][/SIZE] [COLOR=blue][SIZE=4][FONT=.VnTime] Trén CuO víi mét oxit kim lo¹i ®¬n hãa trÞ II theo tû lÖ mol 1:2 ®îc hçn hîp A. DÉn mét luång khÝ H2 d ®i qua 3,6 gam A nung nãng thu ®îc hçn hîp B. §Ó hßa tan hÕt B cÇn 60 ml dung dÞch HNO3 nång ®é 2,5M vµ thu ®îc V lÝt khÝ NO duy nhÊt(®ktc) vµ dung dÞch chØ chøa nitat kimlo¹i. X¸c ®Þnh kim lo¹i hãa trÞ II nãi trªn vµ tÝnh V.[/FONT][/SIZE][/COLOR] [SIZE=4][FONT=.VnTime] [/FONT][FONT=.VnArialH]Gi¶i: [/FONT][FONT=.VnTime]Gäi oxit kim lo¹i ph¶i t×m lµ MO vµ a vµ 2a lµ sè mol CuO vµ MO trong A.[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]V× hidro chØ khö ®îc nh÷ng oxit kim lo¹i ®øng sau nh«m trong d·y ®iÖn hãa nªn cã 2 kh¶ n¨ng x¶y ra:[/SIZE][/FONT] [FONT=.VnTime][SIZE=4]* Trêng hîp 1: M ®øng sau nh«m trong d·y ®iÖn hãa[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] CuO + H2 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Cu + H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] MO + H2 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] M + H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 3Cu + 8HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 3Cu(NO3)2 + 2NO [/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime]+ 4H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 3M + 8HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 3 M(NO3)2 + 2NO [/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime]+ 4H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime]Ta cã hÖ pt: Gi¶i hÖ pt cho a = 0,01875 vµ M = 40 [/FONT][FONT=Symbol][FONT=Symbol]~[/FONT][/FONT][FONT=.VnTime] Ca[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Trêng hîp nµy lo¹i v× Ca ®øng tríc Al trong d·y thÕ ®iÖn hãa.[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]* Trêng hîp 2: M ®øng tríc nh«m trong d·y ®iÖn hãa[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] CuO + H2 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Cu + H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 3Cu + 8HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 3Cu(NO3)2 + 2NO [/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime]+ 4H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] MO + 2HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] M(NO3)2 + 2H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime]Ta cã hÖ pt: Gi¶i hÖ pt cho a = 0,01875 vµ M = 24 [/FONT][FONT=Symbol][FONT=Symbol]~[/FONT][/FONT][FONT=.VnTime] Mg[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]NghiÖm nµy hîp lý vµ V= . 22,4 = 0,28 lÝt. [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] [/FONT][/SIZE] [SIZE=4][FONT=.VnTime][B][U][COLOR=blue]Bµi 5.[/COLOR][/U][/B][U][COLOR=blue] [/COLOR][/U][/FONT][/SIZE] [COLOR=blue][SIZE=4][FONT=.VnTime] X,Y lµ kim lo¹i ®¬n hãa trÞ II vµ III . Hßa tan hÕt 14,0 gam hçn hîp X,Y b»ng axit HNO3 tho¸t ra 14,784 lÝt (27,30C vµ 1,1atm) hçn hîp 2 khÝ oxit cã mµu n©u vµ cã tû khèi so víi He = 9,56 , dung dÞch nhËn ®îc chØ chøa nitrat kim lo¹i. Cïng lîng hçn hîp 2 kim lo¹i trªn cho t¸c dông víi axit HCl d th× còng tho¸t ra 14,784 lÝt khÝ (27,30C vµ 1atm) vµ cßn l¹i 3,2 gam chÊt r¾n kh«ng tan.[/FONT][/SIZE][/COLOR] [COLOR=blue][FONT=.VnTime][SIZE=4]X¸c ®Þnh X, Y vµ tÝnh % lîng mçi kim lo¹i trong hçn hîp ®Çu.[/SIZE][/FONT][/COLOR] [SIZE=4][FONT=.VnArialH]Gi¶i: [/FONT][FONT=.VnTime]Sè mol khÝ = 0,66 vµ 0,6. [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Tõ MTB= 9,56. 4 = 38,24 suy ra NO2 > 38,24 nªn khÝ cßn l¹i ph¶i lµ NO = 30 < 38,24[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Vµ tÝnh ®îc NO = 0,32 mol vµ NO2 = 0,34 mol[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 3X + 8HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 3X(NO3)2 + 2NO [/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime]+ 4H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Y + 4HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Y(NO3)3 + NO [/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime]+ 2H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] X + 4HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] X(NO3)2 + 2NO[/FONT][FONT=.VnTime]2 [/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime]+ 2H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Y + 6HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Y(NO3)3 + 3NO[/FONT][FONT=.VnTime]2 [/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime]+ 3H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] X + 2HCl [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] XCl2 + H[/FONT][FONT=.VnTime]2 [/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] hoÆc 2Y + 6HCl [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 2YCl3 + 3H[/FONT][FONT=.VnTime]2 [/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] [/FONT][/SIZE] [FONT=.VnTime][SIZE=4]BiÖn luËn: * NÕu kim lo¹i Y kh«ng tan trong axit HCl[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] Theo pt: sè mol X = 0,6 vµ lîng X = 10,8 gam nªn X = = 18 (kh«ng tháa m·n kim lo¹i nµo)[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]* VËy kim lo¹i X kh«ng tan trong axit HCl [/SIZE][/FONT] [SIZE=4][FONT=.VnTime] Theo pt: sè mol Y = 0,4 vµ lîng Y = 14- 3,2= 10,8 gam nªn Y = = 27 [/FONT][FONT=Symbol][FONT=Symbol]~[/FONT][/FONT][FONT=.VnTime] Al [/FONT][/SIZE] [FONT=.VnTime][SIZE=4]§Æt sè mol X b»ng a: tæng sè e nhêng = 0,4. 3 + 2a = 1,2 + 2a [/SIZE][/FONT] [SIZE=4][FONT=.VnTime] tæng sè e thu = 0,32. 3 + 0,34 = 1,30 [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Theo qui t¾c b¶o toµn sè mol e: 1,2 + 2a = 1,3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] a = 0,05[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] VËy X = = 64 [/FONT][FONT=Symbol][FONT=Symbol]~[/FONT][/FONT][FONT=.VnTime] Cu vµ % Al = 77,14% ; %Cu = 22,86%[/FONT][/SIZE] [SIZE=4][FONT=.VnTime][B][U][COLOR=blue]Bµi 6.[/COLOR][/U][/B][U][COLOR=blue] [/COLOR][/U][/FONT][/SIZE] [COLOR=blue][SIZE=4][FONT=.VnTime] Cho 21,52 gam hçn hîp A gåm kim lo¹i X ®¬n hãa trÞ II vµ muèi nitrat cña nã vµo b×nh kÝn dung tÝch kh«ng ®æi 3 lÝt (kh«ng chøa kh«ng khÝ) råi nung b×nh ®Õn nhiÖt ®é cao ®Ó ph¶n øng x¶y ra hoµn toµn, s¶n phÈm thu ®îc lµ oxit kim lo¹i. Sau ph¶n øng ®a b×nh vÒ 54,60C th× ¸p suÊt trong b×nh lµ P. Chia ®«i chÊt r¾n trong b×nh sau ph¶n øng: phÇn 1 ph¶n øng võa ®ñ víi 667ml dung dÞch HNO3 nång ®é 0,38M tho¸t ra khÝ NO duy nhÊt vµ dung dÞch chØ chøa nitrat kim lo¹i. PhÇn 2 ph¶n øng võa hÕt víi 300ml dung dÞch H2SO4 lo·ng 0,2M thu ®îc dung dÞch B.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime] a) X¸c ®Þnh kim lo¹i X vµ tÝnh % lîng mçi chÊt trong A.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime] b) TÝnh P.[/FONT][/SIZE][/COLOR] [SIZE=4][FONT=.VnArialH]Gi¶i: [/FONT][FONT=.VnTime]sè mol HNO3 = 0,38. 0,667 = 0,25346 vµ sè mol H2SO4 = 0,3. 0,2 = 0,06 [/FONT][FONT=.VnArialH] [/FONT][/SIZE] [FONT=.VnTime][SIZE=4]§Æt sè mol X(NO3)2 vµ X ban ®Çu lµ a vµ b.[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] 2X(NO3)2 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 2XO + 4NO2 + O[/FONT][FONT=.VnTime]2 [/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] a a 2a 0,5a[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 2X + O2 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 2XO do ph¶n øng víi HNO3 cã khÝ NO[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] nªn X cã d vµ [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] a 0,5a a phÇn d = b – a (mol)[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] XO + 2HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] X(NO3)2 + H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 3X + 8HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 3X(NO3)2 + 2NO [/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime]+ 4H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] XO + H2SO4 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] XSO4 + H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] X + H2SO4 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] XSO4 + H[/FONT][FONT=.VnTime]2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Theo pt: sè mol HNO3 ph¶n øng = 2a + (b - a) = 0,253 hay a + 2b = 0,38 (1)[/FONT][/SIZE] [SIZE=4][FONT=.VnTime][U]BiÖn luËn:[/U] * NÕu M ®øng tríc hidro trong d·y ®iÖn hãa th× theo pt[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] a + (b - a) = 0,06 hay a + b = 0,12 (2). Gi¶i (1)(2) cho a = – 0,14 < 0 (lo¹i)[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]* VËy M ®øng sau hidro trong d·y ®iÖn hãa vµ kh«ng t¸c dông víi H2SO4 lo·ng,[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] khi ®ã a = 0,06 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] b = 0,16 vµ 0,06(M + 124) + 0,16M = 21,52 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] M = 64 [/FONT][FONT=Symbol][FONT=Symbol]~[/FONT][/FONT][FONT=.VnTime] Cu[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]Suy ra % Cu = 47,5 % vµ % Cu(NO3)2 = 52,5%[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] Sau khi nung trong b×nh chøa 0,12 mol NO2 nªn P = = 1,07 atm.[/FONT][/SIZE] [SIZE=4][FONT=.VnTime][B][U][COLOR=blue]Bµi 7.[/COLOR][/U][/B][U][COLOR=blue] [/COLOR][/U][/FONT][/SIZE] [COLOR=blue][SIZE=4][FONT=.VnTime] Hßa tan hoµn toµn 2,52 gam hçn hîp Mg vµ Al b»ng dung dÞch HCl thu ®îc 2,688 lÝt hidro (®ktc). Còng lîng hçn hîp nµy nÕu hßa tan hoµn toµn b»ng H2SO4 ®Æc nãng th× thu ®îc 0,03 mol mét s¶n phÈm duy nhÊt h×nh thµnh do sù khö S+6. [/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime] a) X¸c ®Þnh s¶n phÈm duy nhÊt nãi trªn.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime] b) NÕu hßa tan hoµn toµn còng lîng hçn hîp trªn b»ng dung dÞch HNO3 10,5%[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime](d =1,2 g/ml) th× thu ®îc 0,03 mol mét s¶n phÈm duy nhÊt h×nh thµnh do sù khö N+5.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime]TÝnh thÓ tÝch tèi thiÓu dung dÞch HNO3 ®· dïng.[/FONT][/SIZE][/COLOR] [SIZE=4][FONT=.VnArialH]Gi¶i: [/FONT][FONT=.VnTime]a)[/FONT][FONT=.VnArialH] [/FONT][FONT=.VnTime]Mg + 2HCl [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] MgCl2 + H[/FONT][FONT=.VnTime]2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 2Al + 6HCl [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 2AlCl3 + 3H[/FONT][FONT=.VnTime]2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][/SIZE] [FONT=.VnTime][SIZE=4]LËp hÖ pt vµ gi¶i hÖ pt cho Mg = 0,06 mol vµ Al = 0,04 mol[/SIZE][/FONT] [SIZE=4][FONT=.VnArialH] [/FONT][FONT=.VnTime]§Æt sè oxihãa cña lu huúnh trong s¶n phÈm lµ x ta cã:[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Al – 3e [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Al3+[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Mg – 2e [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Mg2+ tæng sè mol e nhêng = 0,04. 3 + 0,06. 2 = 0,24[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] S6+ + (6 – xe) [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] S x tæng sè mol e thu = (6 - x). 0,03 [/FONT][/SIZE] [SIZE=4][FONT=.VnTime]Theo quy t¾c b¶o toµn sè mol e: (6 - x). 0,03 = 0,24 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] x = – 2 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] s¶n phÈm lµ H2S[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]b) N5+ + (5 – ye) [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] S y t¬ng tù: (5 - x). 0,03 = 0,24 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] x = – 3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] s¶n phÈm lµ NH3[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 4Mg +10 HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 4Mg(NO3)2 + NH4NO3 + 3H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 8Al +30 HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 8Al(NO3)3 + 3NH4NO3 + 9H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]Theo pt: sè mol HNO3 ph¶n øng = 0,06. 2,5 + 0,04. 3,75 = 0,3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] V[/FONT][/SIZE][FONT=.VnTime]HNO3 = [SIZE=4]150 ml.[/SIZE][/FONT] [SIZE=4][FONT=.VnTime][B][U][COLOR=blue]Bµi 8.[/COLOR][/U][/B][U][COLOR=blue] [/COLOR][/U][/FONT][/SIZE] [COLOR=blue][SIZE=4][FONT=.VnTime] Mét miÕng Mg bÞ oxihãa mét phÇn ®îc chia lµm 2 phÇn b»ngnhau: [/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime]- PhÇn 1 cho hßa tan hÕt trong dung dÞch HCl th× tho¸t ra 3,136 lÝt khÝ. C« c¹n dung dÞch thu ®îc 14,25 gam chÊt r¾n A.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime]- PhÇn 2 cho hßa tan hÕt trong dung dÞch HNO3 th× tho¸t ra 0,448 lÝt khÝ X nguyªn chÊt. C« c¹n dung dÞch thu ®îc 23 gam chÊt r¾n B.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime]a) TÝnh % sè mol Mg ®· bÞ oxihãa.(c¸c thÓ tÝch khÝ ®Òu ®o ë ®ktc)[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime]b) X¸c ®Þnh khÝ X. [/FONT][/SIZE][/COLOR] [SIZE=4][FONT=.VnArialH]Gi¶i: [/FONT][FONT=.VnTime]a)[/FONT][FONT=.VnArialH] [/FONT][FONT=.VnTime]sè[/FONT][FONT=.VnArialH] [/FONT][FONT=.VnTime]mol[/FONT][FONT=.VnArialH] [/FONT][FONT=.VnTime]khÝ H2 = 0,14 vµ sè mol khÝ X = 0,02[/FONT][FONT=.VnArialH] [/FONT][/SIZE] [SIZE=4][FONT=.VnArialH] [/FONT][FONT=.VnTime]2Mg + O[/FONT][FONT=.VnTime]2 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 2MgO[/FONT][FONT=.VnArialH] [/FONT][/SIZE] [SIZE=4][FONT=.VnArialH] [/FONT][FONT=.VnTime]Mg + 2HCl [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] MgCl2 + H[/FONT][FONT=.VnTime]2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][/SIZE] [SIZE=4][FONT=.VnTime] MgO + 2HCl [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] MgCl2 + H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]Theo pt: Mg cha bÞ oxihãa = 0,14 mol [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] MgO = – 0,14 = 0,01 mol[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] % Mg bÞ oxihãa = . 100% = 6,67%[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]b) Mg + HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Mg(NO3)2 + X + H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] MgO + 2HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Mg(NO3)2 + 2H2O[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]BiÖn luËn: Theo tÝnh to¸n trªn sè mol Mg(NO3)2 = 0,15 [/SIZE][/FONT] [SIZE=4][FONT=.VnTime] nªn lîng Mg(NO3)2 = 0,15. 148 = 22,2 gam < 23[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]VËy trong B kh«ng chØ cã Mg(NO3)2 mµ cßn ph¶i cã NH4NO3 = = 0,01 mol[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] 4Mg + 10HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 4Mg(NO3)2 + NH4NO3 + 3H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 0,04 0,01[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] (5x-2y)Mg + (12x-4y)HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] (5x-2y)Mg(NO3)2 + 2N[/FONT][/SIZE][FONT=.VnTime]x[SIZE=4]O[/SIZE]y[SIZE=4] [/SIZE][/FONT][SIZE=4][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] + (6x-2y)H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 0,14 – 0,04 = 0,1 0,02[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Theo pt: = [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 5x – 2y = 10 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] x = 2; y = 0 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] khÝ X lµ N2.[/FONT][/SIZE] [COLOR=blue][SIZE=4][FONT=.VnTime] [B][U]Bµi 9[/U][/B]:[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime] Cho 23,52g hçn hîp 3 kim lo¹i Mg, Fe, Cu vµo 200ml dung dÞch HNO3 3,4M khuÊy ®Òu thÊy tho¸t ra mét khÝ duy nhÊt h¬i nÆng h¬n kh«ng khÝ, trong dung dÞch cßn d mét kim lo¹i cha tan hÕt, ®æ tiÐp tõ tõ dung dÞch H2SO4 5M vµo, chÊt khÝ trªn l¹i tho¸t ra cho dÕn khi kim lo¹i võa tan hÕt th× mÊt ®óng 44ml, thu ®îc dd A. LÊy 1/2 dd A, cho dd NaOH cho ®Õn d vµo, läc kÕt tña, röa råi nung ngoµi kh«ng khÝ ®Õn khèi lîng kh«ng ®æi thu ®îc chÊt r¾n B nÆng 15,6g.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime] 1-TÝnh % sè mol mçi kim lo¹i trong hçn hîp.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime] 2-TÝnh nång ®é c¸c ion (trõ ion H+‑, OH-) trong dung dÞch A.[/FONT][/SIZE][/COLOR] [SIZE=4][FONT=.VnArialH]Gi¶i: [/FONT][FONT=.VnTime]Gäi x, y, z lµ sè mol Mg, Fe, Cu trong hçn hîp, ta cã : [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 24x + 56y + 64z = 23,52 [/FONT][FONT=Symbol][FONT=Symbol]«[/FONT][/FONT][FONT=.VnTime] 3x + 7y + 8z = 2,94 (a) [/FONT][/SIZE] [FONT=.VnTime][SIZE=4]§ång cßn d cã c¸c ph¶n øng:[/SIZE][/FONT] [FONT=.VnTime][SIZE=4]Cho e: NhËn e:[/SIZE][/FONT] [FONT=.VnTime][SIZE=4]Mg - 2e = Mg2+ (1) NO3- + 3e + 4H+ = NO + 2H2O (4)[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] Fe - 3e = Fe3+ (2) Cu + Fe3+ = Cu2+ + Fe2+ (5)[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]Cu - 2e = Cu2+ (3)[/SIZE][/FONT] [FONT=.VnTime][SIZE=4]Ph¬ng tr×nh ph¶n øng hoµ tan Cu d:[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] 3Cu + 4H2SO4 + 2NO3- = 3CuSO4 + SO42- + 2NO + H2O (6)[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]Tõ Pt (6) tÝnh ®îc sè mol Cu d: = = 0,165 mol[/SIZE][/FONT] [FONT=.VnTime][SIZE=4]Theo c¸c ph¬ng tr×nh (1), (2), (3), (4), (5): sè mol cho b»ng sè mol e nhËn:[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] 2(x + y + z – 0,165) = [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime]3,4.0,2 – 2(x + y + z – 0,165)[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime].3[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] x + y + z = 0,255 + 0,165 = 0,42 (b)[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]Tõ khèi lîng c¸c oxit MgO; Fe2O3; CuO, cã ph¬ng tr×nh: .40 + .160 + . 80 = 15,6 (c)[/SIZE][/FONT] [FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4]HÖ ph¬ng tr×nh rót ra tõ (a), (b), (c): 3x + 7y + 8z = 2,94[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] x + y + z = 0,42[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] x + 2y + 2z = 0,78[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]Gi¶i ®îc: x = 0,06; y = 0,12; z = 0,24.[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] [/FONT][FONT=Symbol][FONT=Symbol]%[/FONT][/FONT][FONT=.VnTime] lîng Mg = 6,12[/FONT][FONT=Symbol][FONT=Symbol]%[/FONT][/FONT][FONT=.VnTime] ; [/FONT][FONT=Symbol][FONT=Symbol]%[/FONT][/FONT][FONT=.VnTime] lîng Fe = 28,57[/FONT][FONT=Symbol][FONT=Symbol]%[/FONT][/FONT][FONT=.VnTime] ; [/FONT][FONT=Symbol][FONT=Symbol]%[/FONT][/FONT][FONT=.VnTime] lîng Cu = 65,31[/FONT][FONT=Symbol][FONT=Symbol]%[/FONT][/FONT][/SIZE] [SIZE=4][FONT=.VnTime]2/ TÝnh nång ®é c¸c ion trong dd A (trõ H+, OH-) [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime]Mg2+[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] = = 0,246 M[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime]Cu2+[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] = 0,984 M ; [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime]Fe2+[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] = 0,492 M ; [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime]SO[/FONT][FONT=.VnTime]42-[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] = 0,9 M ; [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime]NO[/FONT][FONT=.VnTime]3-[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] = 1,64 M[/FONT][/SIZE] [COLOR=blue][SIZE=4][FONT=.VnTime] [B][U]Bµi 10[/U][/B]:[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime] Hoµ tan hoµn toµn hçn hîp FeS vµ FeCO3 b»ng dung dÞch HNO3 ®Æc, nãng thu ®îc hçn hîp (B) gåm 2 khÝ X vµ Y cã tû khèi ®èi víi H2 lµ 22,8 vµ cßn dung dÞch (A) cã pH < 3.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime]a)TÝnh tû lÖ sè mol c¸c muèi Fe2+ trong hçn hîp ban ®Çu.[/FONT][/SIZE][/COLOR] [SIZE=4][COLOR=blue][FONT=.VnTime]b)Lµm l¹nh hçn hîp khÝ (B) xuèng nhiÖt ®é thÊp h¬n ®îc hçn hîp (B[/FONT][/COLOR][COLOR=blue][FONT=Symbol][FONT=Symbol]¢[/FONT][/FONT][/COLOR][COLOR=blue][FONT=.VnTime]) gåm 3 khÝ X,Y,Z cã tû khèi so víi H2 b»ng 28,5. TÝnh phÇn tr¨m theo thÓ tÝch cña hçn hîp khÝ (B[/FONT][/COLOR][COLOR=blue][FONT=Symbol][FONT=Symbol]¢[/FONT][/FONT][/COLOR][COLOR=blue][FONT=.VnTime]).[/FONT][/COLOR][/SIZE] [SIZE=4][COLOR=blue][FONT=.VnTime]c) [/FONT][/COLOR][COLOR=blue][FONT=.VnTimeH]ë[/FONT][/COLOR][COLOR=blue][FONT=.VnTime] -11oC hçn hîp (B[/FONT][/COLOR][COLOR=blue][FONT=Symbol][FONT=Symbol]¢[/FONT][/FONT][/COLOR][COLOR=blue][FONT=.VnTime]) chuyÓn sang (B[/FONT][/COLOR][COLOR=blue][FONT=Symbol][FONT=Symbol]²[/FONT][/FONT][/COLOR][COLOR=blue][FONT=.VnTime]) gåm 2 khÝ. TÝnh tû khèi cña (B[/FONT][/COLOR][COLOR=blue][FONT=Symbol][FONT=Symbol]²[/FONT][/FONT][/COLOR][COLOR=blue][FONT=.VnTime]) so víi H2.[/FONT][/COLOR][/SIZE] [SIZE=4][FONT=.VnArialH]Gi¶i:[/FONT][FONT=.VnTime] a) PT pø: FeS + 12HNO[/FONT][FONT=.VnTime]3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Fe(NO3)3 + H2SO4 + 9NO[/FONT][FONT=.VnTime]2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] + 5H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] FeCO3 + 4HNO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Fe(NO3)3 + CO2 + NO[/FONT][FONT=.VnTime]2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] + 2H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]§Æt: nFeS = a mol , nFeCO3 = b mol [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] nNO2 = 9a + b vµ nCO2 = b [/FONT][/SIZE] [SIZE=4][FONT=.VnTime]Ta cã : = 22,8 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] a:b = 1:3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] n FeS : n FeCO3 = 1: 3 (b = 3a)[/FONT][/SIZE] [FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4]b) Lµm l¹nh B cã ph¶n øng : 2NO2 N2O4 khi ®ã M (N2O4) = 92 lµm t¨ng vµ = 57[/SIZE][/FONT] [SIZE=4][FONT=.VnTime]Gäi x lµ sè mol N2O4 trong hçn hîp B, vËy B[/FONT][FONT=Symbol][FONT=Symbol]¢[/FONT][/FONT][FONT=.VnTime] gåm:[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]NO2 = (9a + b) – 2x = 4b - 2x ; N2O4 = x vµ CO2 = b [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] = 57[/FONT][/SIZE] [SIZE=4][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] b = x [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Tæng B’ = 4b gåm NO2 = 2b [/FONT][FONT=Symbol][FONT=Symbol]~[/FONT][/FONT][FONT=.VnTime] 50[/FONT][FONT=Symbol][FONT=Symbol]%[/FONT][/FONT][FONT=.VnTime] ; N2O4 = b [/FONT][FONT=Symbol][FONT=Symbol]~[/FONT][/FONT][FONT=.VnTime] 25[/FONT][FONT=Symbol][FONT=Symbol]%[/FONT][/FONT][FONT=.VnTime] ; CO2 = b [/FONT][FONT=Symbol][FONT=Symbol]~[/FONT][/FONT][FONT=.VnTime] 25[/FONT][FONT=Symbol][FONT=Symbol]%[/FONT][/FONT][/SIZE] [SIZE=4][FONT=.VnTime]c) ë – 110C ph¶n øng dime ho¸ x¶y ra hoµn toµn, B[/FONT][FONT=Symbol][FONT=Symbol]¢[FONT=Symbol]¢[/FONT][/FONT][/FONT][FONT=.VnTime] gåm N2O4 = 2b vµ CO2 = b[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] tØ khèi so víi hidro = = 38[/FONT][/SIZE] [FONT=.VnTime][SIZE=4] [/SIZE][/FONT] [SIZE=4][FONT=.VnTime][B][U][COLOR=blue]Bµi 11[/COLOR][/U][/B][COLOR=blue]:[/COLOR][/FONT][/SIZE] [COLOR=blue][SIZE=4][FONT=.VnTime] Cho tõ tõ khÝ CO qua èng chøa 6,4gam CuO ®un nãng. KhÝ ra khái èng ®îc hÊp thô hoµn toµn b»ng 150ml dung dÞch níc v«i trong nång ®é 0,1M thÊy t¸ch ra 1,0 gam kÕt tña tr¾ng, ®un s«i phÇn níc läc l¹i thÊy cã vÈn ®ôc. ChÊt r¾n cßn l¹i trong èng ®îc cho vµo 500ml dung dÞch HNO3 0,32M tho¸t ra V1 lÝt khÝ NO2 nÕu thªm 760ml dung dÞch HCl 1,333M vµo dung dÞch sau ph¶n øng th× l¹i tho¸t ra thªm V2 lÝt khÝ NO n÷a. NÕu tiÕp tôc thªm 24 gam Mg th× thÊy tho¸t ra V3 lÝt khÝ hçn hîp khÝ N2 vµ H2, läc dung dÞch cuèi cïng thu ®îc chÊt r¾n X.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime]a/ ViÕt ph¬ng tr×nh ph¶n øng vµ tÝnh V1,V2,V3(®ktc).[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime]b/ TÝnh thµnh phÇn X( gi¶ thiÕt c¸c ph¶n øng x¶y ra hoµn toµn).[/FONT][/SIZE][/COLOR] [SIZE=4][FONT=.VnArialH]Gi¶i:[/FONT][FONT=.VnTime]a)[/FONT][FONT=.VnArialH] [/FONT][FONT=.VnTime] CO + CuO [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Cu + CO2 sè mol CO2 = 0,02[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] CO2 + Ca(OH)2 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] CaCO3 + H2O sè mol Cu = 0,02[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 2CO2 + Ca(OH)2 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Ca(HCO3)2 sè mol CuO = 0,06[B][U][COLOR=blue][/COLOR][/U][/B][/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Ca(HCO3)2 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] CaCO3 + CO[/FONT][FONT=.VnTime]2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime]+ H2O sè mol HNO3 = 0,16[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] CuO + 2H+ [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Cu2+ + H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 3Cu + 8H+ + 2NO3– [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 3Cu2+ + 2NO[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] + 4H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 0,015 0,04 0,01[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]Theo pt: V1 = 0,01. 22,4 = 0,224 lÝt ; V2 = V1 = 0,07467 lÝt[/SIZE][/FONT] [FONT=.VnTime][SIZE=4]Khi thªm mol HCl, ph¶n øng l¹i tiÕp tôc x¶y ra[/SIZE][/FONT] [SIZE=4][FONT=.VnTime]Thªm 1 mol Mg: 5Mg + 12H+ + 2NO3– [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 5Mg2+ + N[/FONT][FONT=.VnTime]2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] + 6H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Mg + Cu2+ [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Mg2+ + Cu[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Mg + 2H+ [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Mg2+ + H[/FONT][FONT=.VnTime]2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] V3 = V[/FONT][/SIZE][FONT=.VnTime]N2[SIZE=4] + V[/SIZE]H2 [/FONT] [SIZE=4][FONT=.VnTime]mµ sè mol N2 = (0,16 - ) = vµ H2 = [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] nªn V3 = 22,4 = 2,9867 lÝt[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]b) Suy ra Cu = 0,08. 64 = 5,12 gam chiÕm 30,19% [/SIZE][/FONT] [SIZE=4][FONT=.VnTime] vµ Mg = .24 = 11,84 gam chiÕm 69,81%[/FONT][/SIZE] [SIZE=4][FONT=.VnTime][B][U][COLOR=blue]Bµi 12[/COLOR][/U][/B][COLOR=blue]:[/COLOR][/FONT][/SIZE] [COLOR=blue][SIZE=4][FONT=.VnTime]1/Hoµ tan khÝ SO2 vµo H2O cã c¸c c©n b»ng sau:[/FONT][/SIZE][/COLOR] [FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][COLOR=blue][SIZE=4][FONT=.VnTime] SO2 + H2O H2SO3 (1)[/FONT][/SIZE][/COLOR] [FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][COLOR=blue][SIZE=4][FONT=.VnTime] H2SO3 H+ + HSO3– (2)[/FONT][/SIZE][/COLOR] [FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][COLOR=blue][SIZE=4][FONT=.VnTime] HSO3 H+ + SO32– (3)[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime]Nång ®é SO2 thay ®æi nh thÕ nµo? (gi¶i thÝch) khi lÇn lît t¸c ®éng nh÷ng yÕu tè sau:[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime] a)§un nãng dung dÞch.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime] b)Thªm dung dÞch HCl.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime] c)Thªm dung dÞch NaOH.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime] d)Thªm dung dÞch KMnO4.[/FONT][/SIZE][/COLOR] [SIZE=4][FONT=.VnArialH]Gi¶i: [/FONT][FONT=.VnTime] a)§un nãng dd: SO2 tho¸t khái dd [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime] SO[/FONT][FONT=.VnTime]2 [/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] trong dung dÞch gi¶m.[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] b)Thªm dd HCl: Cb»ng (2) (3) chuyÓn sang tr¸i [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime]C©n b»ng (1) chuyÓn sang tr¸i[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime] H2SO[/FONT][FONT=.VnTime]3 [/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] t¨ng [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime] SO[/FONT][FONT=.VnTime]2 [/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] t¨ng.[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] c)Thªm NaOH: NaOH + SO2 = NaHSO3 [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 2 NaOH + SO2 = Na2SO3 + H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime]Cb»ng (2) (3) chuyÓn sang ph¶i [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime]C©n b»ng (1) chuyÓn sang ph¶i [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime]SO[/FONT][FONT=.VnTime]2 [/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] gi¶m[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] d)Thªm KMnO4. Cã pø: [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] lµm gi¶m [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime] SO[/FONT][FONT=.VnTime]2 [/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] .[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 5SO2 + 2KMnO4 + 2H2O = 2H2SO4 + 2MnSO4 + K2SO4[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] SO32– + 2KMnO4 + 2H2O = H2SO4 + K2MnO4 + MnO42– [/FONT][/SIZE] [B][U][COLOR=blue][SIZE=4][FONT=.VnTime]Bµi 13:[/FONT][/SIZE][/COLOR][/U][/B] [COLOR=blue][SIZE=4][FONT=.VnTime]1- DÉn ra 5 ph¬ng tr×nh ph¶n øng kh¸c nhau cã t¹o ra khÝ NO2.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime]2-a)TÝnh pH cña dung dÞch HCl nång ®é 0,5.10-7 mol/lÝt.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime] b)TÝnh pH cña dung dÞch X ®îc t¹o thµnh khi trén 200ml dung dÞch HA 0,1M (Ka = 10-3.75) víi 200ml dung dÞch KOH 0.05M; pH cña dung dÞch X thay ®æi nh thÕ nµo khi thªm 10-3 mol HCl vµo dung dÞch X.[/FONT][/SIZE][/COLOR] [FONT=.VnArialH][SIZE=4]Gi¶i:[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] 1/ 2HNO3 =2NO2 + 1/2O2 + 2H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Cu + 4HNO3 = Cu(NO3)2 + 2NO2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] + 2H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 2AgNO3 = 2Ag + O2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] + 2NO[/FONT][FONT=.VnTime]2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][/SIZE] [SIZE=4][FONT=.VnTime] C + 4HNO3 = CO2 + 4NO2[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] + 2H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 2NO + O2 = 2NO2[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 2/a) [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime] H[/FONT][FONT=.VnTime]+[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] . 0,5.10-7 do nång ®é nhá [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] ph¶i tÝnh ®Õn c©n b»ng cña H2O [/FONT][/SIZE] [FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][SIZE=4][FONT=.VnTime] H2O H+ + OH –[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] HCl [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] H+ + Cl –[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Theo ®Þnh luËt b¶o toµn ®iÖn tÝch: [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime] H[/FONT][FONT=.VnTime]+[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] = [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime] Cl[/FONT][FONT=.VnTime]-[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] + [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime]OH[/FONT][FONT=.VnTime]-[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime] H+[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] = 0,5.10-7 + [/FONT][/SIZE] [SIZE=4][FONT=.VnTime] [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime] H+[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime]2 – 0,5.10 [/FONT][FONT=.VnTime]–7[/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime] H[/FONT][FONT=.VnTime]+[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] – 10-14 = 0. Gi¶i ®îc: [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime] H[/FONT][FONT=.VnTime]+[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] = 1,28.10[/FONT][FONT=.VnTime]-7 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] pH [/FONT][FONT=Symbol][FONT=Symbol]»[/FONT][/FONT][FONT=.VnTime] 6,9[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]b) nHA = 0,1.0,2 = 0,02 mol ; nKOH = 0,05.0,2 = 0,01 mol[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] KOH + HA [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] KA + H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 0,01 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 0,01[/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 0,01 Theo ph¬ng tr×nh HA cßn d = 0,01 mol[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]Trong d2 X: CHA = CKA = = 0,025M. XÐt c¸c c©n b»ng sau: [/SIZE][/FONT] [FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][SIZE=4][FONT=.VnTime] H2O H+ + OH- KW = 10-14 (1) [/FONT][/SIZE] [FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][SIZE=4][FONT=.VnTime] HA H+ + A- KHA = 10-375 (2)[/FONT][/SIZE] [FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][SIZE=4][FONT=.VnTime] A- + H2O HA + OH- KB = KHA-1. KW = 10-10,25 (3) [/FONT][/SIZE] [SIZE=4][FONT=.VnTime]So s¸nh (1) víi (2) [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] KHA >> K[/FONT][FONT=.VnTime]W [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] bá qua (1) [/FONT][/SIZE] [SIZE=4][FONT=.VnTime]So s¸nh (2) víi (3) [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] KHA >> K[/FONT][FONT=.VnTime]B [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] bá qua(3) [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] Dung dÞch X lµ dung dÞch ®Öm axit[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]cã pH = pKa + lg = 3,75 + lg = 3,75[/SIZE][/FONT] [SIZE=4][FONT=.VnTime] [/FONT][FONT=Symbol][FONT=Symbol]*[/FONT][/FONT][FONT=.VnTime] Khi thªm 10-3 mol HCl: KA + HCl [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] KCl + HA[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] 0,001 [/FONT][FONT=Symbol][FONT=Symbol]¬[/FONT][/FONT][FONT=.VnTime] 0,001 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] 0,001 (mol)[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime]HA[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] = = 0,0275 M vµ [/FONT][FONT=Symbol][FONT=Symbol][[/FONT][/FONT][FONT=.VnTime]KA[/FONT][FONT=Symbol][FONT=Symbol]][/FONT][/FONT][FONT=.VnTime] = = 0,0225M .[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]Dung dÞch thu ®îc vÉn lµ dung dÞch ®Öm axit nªn pH = 3,75 + lg = 3,66[/SIZE][/FONT] [SIZE=4][FONT=.VnTime][B][U][COLOR=blue]Bµi 14:[/COLOR][/U][/B][U][COLOR=blue] [/COLOR][/U][/FONT][/SIZE] [COLOR=blue][SIZE=4][FONT=.VnTime] 1/ Dung dÞch A gåm c¸c cation: NH4+ ; Na+ ; Ba2+ vµ 1 anion X cã thÓ lµ mét trong c¸c anion sau: CH3COO – ; NO3–; SO42– ; CO32– ; PO43– . [I] [/I]Hái X lµ anion nµo? BiÕt r»ng dung dÞch A cã pH = 5 .[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime] 2/ Thªm NaOH d vµo dung dÞch CuSO4 ,thªm tiÕp NH4NO3 vµo dung dÞch ®Õn d cã hiÖn tîng g× x¶y ra? ViÕt ph¬ng tr×nh ph¶n øng.[/FONT][/SIZE][/COLOR] [COLOR=blue][SIZE=4][FONT=.VnTime] 3/ Cho c¸c dung dÞch sau: Al(NO3)3 ; ZnCl2 ; Na2CO3 ; MgSO4 ; NH4NO3 . Kh«ng dïng thªm thuèc thö h·y nªu c¸ch nhËn ra tõng dung dÞch, viÕt ph¬ng tr×nh ph¶n øng d¹ng ion.[/FONT][/SIZE][/COLOR] [SIZE=4][FONT=.VnArialH]Gi¶i: [/FONT][FONT=.VnTime]1/ X lµ NO[/FONT][/SIZE][FONT=.VnTime]3[SIZE=4]– v× NH4NO3: m«i trêng axit pH< 7. [/SIZE][/FONT] [SIZE=4][FONT=.VnTime]2/ + cã kÕt tña mµu xanh: Cu2+ + 2OH- = Cu(OH)2[/FONT][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][/SIZE] [SIZE=4][FONT=.VnTime] + cã khÝ mïi khai : NH4+ + OH- = NH3[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] + H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] + kÕt tña tan t¹o dung dÞch xanh thÉm: Cu(OH)2 + 4NH3 = [Cu(NH3)4]2+ + 2OH-[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]3/ LÊy mçi dung dÞch mét Ýt cho lÉn lît vµo c¸c dung dÞch cßn l¹i hiÖn tîng ®îc ghi trong b¶ng sau:[/SIZE][/FONT] [FONT=.VnTime][SIZE=4] [/SIZE][/FONT] [SIZE=4][FONT=.VnTime]Al(NO3)3[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]ZnCl2[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]Na2CO3[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]MgSO4[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]NH4NO3[/FONT][/SIZE] [FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4][/SIZE][/FONT][FONT=.VnTime][SIZE=4]Al(NO3)3[/SIZE][/FONT] [FONT=.VnTime][SIZE=4] [/SIZE][/FONT] [FONT=.VnTime][SIZE=4]-[/SIZE][/FONT] [SIZE=4][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime], [/FONT][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime]tan (1)[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]-[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]-[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]ZnCl2[/SIZE][/FONT] [FONT=.VnTime][SIZE=4]-[/SIZE][/FONT] [FONT=.VnTime][SIZE=4] [/SIZE][/FONT] [SIZE=4][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime], [/FONT][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime]tan (2)[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]-[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]-[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]Na2CO3[/SIZE][/FONT] [SIZE=4][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime], [/FONT][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime]tan[/FONT][/SIZE] [SIZE=4][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime], [/FONT][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime]tan[/FONT][/SIZE] [FONT=.VnTime][SIZE=4] [/SIZE][/FONT] [SIZE=4][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime]tr¾ng[/FONT][/SIZE] [SIZE=4][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime](khai)[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]MgSO4[/SIZE][/FONT] [FONT=.VnTime][SIZE=4]-[/SIZE][/FONT] [FONT=.VnTime][SIZE=4]-[/SIZE][/FONT] [SIZE=4][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime], [/FONT][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime]tan (3)[/FONT][/SIZE] [FONT=.VnTime][SIZE=4] [/SIZE][/FONT] [SIZE=4][FONT=.VnTime]-[/FONT][/SIZE] [FONT=.VnTime][SIZE=4]NH4NO3[/SIZE][/FONT] [FONT=.VnTime][SIZE=4]-[/SIZE][/FONT] [FONT=.VnTime][SIZE=4]-[/SIZE][/FONT] [SIZE=4][FONT=Symbol][FONT=Symbol][/FONT][/FONT][FONT=.VnTime] (khai)(4)[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]-[/FONT][/SIZE] [FONT=.VnTime][SIZE=4] [/SIZE][/FONT] [SIZE=4][FONT=.VnTime]KÕt luËn[/FONT][/SIZE] [SIZE=4][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime], [/FONT][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime]tan[/FONT][/SIZE] [SIZE=4][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime], [/FONT][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime]tan[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]3[/FONT][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime],[/FONT][FONT=Symbol][FONT=Symbol][/FONT][/FONT][/SIZE] [SIZE=4][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][/SIZE] [SIZE=4][FONT=Symbol][FONT=Symbol][/FONT][/FONT][/SIZE] [SIZE=4][FONT=.VnTime]-NÕu dung dÞch cho 3 trêng hîp kÕt tña vµ 1 trêng hîp cã khÝ mïi khai tho¸t ra lµ dung dÞch Na2CO3:[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Al3+ + 3OH- = Al(OH)3[/FONT][FONT=Symbol][FONT=Symbol]¯[/FONT][/FONT][FONT=.VnTime] (CO32- + H2O [/FONT][FONT=Symbol][FONT=Symbol]Û[/FONT][/FONT][FONT=.VnTime] HCO3- + OH-)[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Al(OH)3 + OH- = AlO2- + 2H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Zn2+ + 2OH- = Zn(OH)2[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Zn(OH)2 + 2OH- =ZnO22- + 2H2O[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Mg2+ + 2OH- = Mg(OH)2 (kh«ng viÕt Mg2+ + CO32- = MgCO3)[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] NH4+ + OH- = NH3 + H2O [/FONT][/SIZE] [SIZE=4][FONT=.VnTime]- NÕu dung dÞch chØ cho 1 kÕt tña vµ kÕt tña kh«ng tan trong thuèc thö d lµ dung dÞch MgSO4.[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]- NÕu dung dÞch chØ cho 1 trêng hîp cã khÝ mïi khai tho¸t ra lµ dung dÞch NH4NO3.[/FONT][/SIZE] [SIZE=4][FONT=.VnTime]- 2 dung dÞch cßn l¹i cã hiÖn tîng gièng nhau lµ Al(NO3)3 vµ ZnCl2[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] Dïng dung dÞch NH4NO3 nhËn ra kÕt tña Zn(OH)2: kÕt tña tan trong dung dÞch NH4NO3 [/FONT][FONT=Symbol][FONT=Symbol]®[/FONT][/FONT][FONT=.VnTime] dung dÞch t¬ng øng lµ ZnCl2 cßn l¹i lµ dung dÞch Al(NO3)3[/FONT][/SIZE] [SIZE=4][FONT=.VnTime] NH4+ [/FONT][FONT=Symbol][FONT=Symbol]Û[/FONT][/FONT][FONT=.VnTime] NH3 + H+ vµ Zn(OH)2 + 4NH3 = [Zn(NH3)4]2+ + 2OH-[B][/B][/FONT][/SIZE] [/QUOTE]

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