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<blockquote data-quote="namlun_1994" data-source="post: 60086" data-attributes="member: 70187"> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; Bai giai ne FeCO3+ 4HNO3→ Fe(NO3)3 +CO2 + NO2 + 2H2O Xmol&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 4x&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; x&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; x&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; x (2) FeS2 +18HNO3→ Fe(NO3)3 + 2H2SO4 +15 NO2 +9 H2O y&nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 18y&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; y&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2y&nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 15y hh khi B gom CO2 va NO2 CO2 44&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 0.4 nCO2
1.425*32= 45.6&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; → ───&nbsp; &nbsp; = 0.25 → x = 0.25*(x +15y)&nbsp; (1)&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; nNO2&nbsp; &nbsp; &nbsp;NO2 46&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 1.6&nbsp; nBa(OH)2= 0.104 (mol)→ nOH- =0.208 mol&nbsp;Fe3+ +3OH- → Fe(OH)3 X +ymol&nbsp; &nbsp; 3(x+y)&nbsp; &nbsp;&nbsp; x+y&nbsp; 2Fe(OH)3→ Fe2O3 + 3H2O x+y&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; (x+y)/2&nbsp; Ba 2+ +SO42- → BaSO4 2y&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 2y&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2y&nbsp; &nbsp;mhhsaupu= 160*(x+y)/2 + 233*2*y = 7.568&nbsp; (2)&nbsp; tu (1)&amp; (2): x= 0.04mol y =0.008 mol mFeCO3= 0.04*116= 4.64g mFeS2 = 0.008*120 =0.96g ta co: CMHNO3 =(C%*d)/(10*MHNO3) = (63*1.44)/(10*63)= 0.144M VHNO3= (4x +18y)/ CM= 0.211 (lit) chi don gian zay thuj chuc pan thi tot&nbsp;</blockquote>

[QUOTE="namlun_1994, post: 60086, member: 70187"] [COLOR=#3e3e3e] [SIZE=4][FONT=Times New Roman] Bai giai ne[/FONT][/SIZE][/COLOR] [COLOR=#3e3e3e] [SIZE=4][FONT=Times New Roman]FeCO3+ 4HNO3→ Fe(NO3)3 +CO2 + NO2 + 2H2O[/FONT][/SIZE][/COLOR] [COLOR=#3e3e3e] [SIZE=4][FONT=Times New Roman]Xmol 4x x x x[/FONT][/SIZE][/COLOR] [COLOR=#3e3e3e] [SIZE=4][FONT=Times New Roman](2) FeS2 +18HNO3→ Fe(NO3)3 + 2H2SO4 +15 NO2 +9 H2O [/FONT][/SIZE][/COLOR] [COLOR=#3e3e3e] [SIZE=4][FONT=Times New Roman]y 18y y 2y 15y[/FONT][/SIZE][/COLOR] [COLOR=#3e3e3e] [SIZE=4][FONT=Times New Roman]hh khi B gom CO2 va NO2[/FONT][/SIZE][/COLOR] [COLOR=#3e3e3e][FONT=Times New Roman] [SIZE=4]CO2 44 0.4[/SIZE][/FONT][/COLOR] [CENTER][CENTER] [SIZE=4][FONT=Times New Roman]nCO2[/FONT][/SIZE][/CENTER][/CENTER] [SIZE=4][FONT=Times New Roman]1.425*32= 45.6 → ─── = 0.25 → x = 0.25*(x +15y) (1)[/FONT][/SIZE] [SIZE=4][FONT=Times New Roman] nNO2[/FONT][/SIZE] [FONT=Times New Roman] [SIZE=4] [/SIZE][/FONT] [FONT=Times New Roman] [SIZE=4] [/SIZE][/FONT] [SIZE=4][FONT=Times New Roman]NO2 46 1.6[/FONT][/SIZE] [FONT=Times New Roman] [SIZE=4] [/SIZE][/FONT] [FONT=Times New Roman] [SIZE=4]nBa(OH)2= 0.104 (mol)→ nOH- =0.208 mol[/SIZE][/FONT] [SIZE=4][FONT=Times New Roman]Fe3+ +3OH- → Fe(OH)3[/FONT][/SIZE] [FONT=Times New Roman] [SIZE=4]X +ymol 3(x+y) x+y[/SIZE][/FONT] [SIZE=4][FONT=Times New Roman] 2Fe(OH)3→ Fe2O3 + 3H2O[/FONT][/SIZE] [FONT=Times New Roman] [SIZE=4]x+y (x+y)/2[/SIZE][/FONT] [FONT=Times New Roman] [SIZE=4] [/SIZE][/FONT] [FONT=Times New Roman] [SIZE=4]Ba 2+ +SO42- → BaSO4[/SIZE][/FONT] [FONT=Times New Roman] [SIZE=4]2y 2y 2y[/SIZE][/FONT] [FONT=Times New Roman] [SIZE=4] [/SIZE][/FONT] [SIZE=4][FONT=Times New Roman]mhhsaupu= 160*(x+y)/2 + 233*2*y = 7.568 (2)[/FONT][/SIZE] [FONT=Times New Roman] [SIZE=4] [/SIZE][/FONT] [FONT=Times New Roman] [SIZE=4]tu (1)& (2): x= 0.04mol y =0.008 mol[/SIZE][/FONT] [FONT=Times New Roman] [SIZE=4]mFeCO3= 0.04*116= 4.64g[/SIZE][/FONT] [FONT=Times New Roman] [SIZE=4]mFeS2 = 0.008*120 =0.96g[/SIZE][/FONT] [FONT=Times New Roman] [SIZE=4]ta co: CMHNO3 =(C%*d)/(10*MHNO3) = (63*1.44)/(10*63)= 0.144M[/SIZE][/FONT] [FONT=Times New Roman] [SIZE=4]VHNO3= (4x +18y)/ CM= 0.211 (lit)[/SIZE][/FONT] [FONT=Times New Roman] [SIZE=4] chi don gian zay thuj chuc pan thi tot[/SIZE][/FONT] [FONT=Times New Roman] [SIZE=4] [/SIZE][/FONT] [/QUOTE]

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