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<blockquote data-quote="quylot1010" data-source="post: 30103" data-attributes="member: 24416">Cau 1: trong lan them dau : OH- ket tua 1 phan Al3+ =&gt; nAl3+ =0,1 mol. &nbsp;&nbsp; trong lan them thu 2:&nbsp; OH- lai lam ket tua them mot phan nua =&gt;nAl3+ =0.04 mol Tong oH- them vao : 0.5 mol&nbsp;Tong oH- lam ket tua Al3+ la 0.42 = 3*0.1+3*0.04 =&gt; 0.42&lt;0.5 &nbsp;=&gt; co 0.08 mol Oh- lam tan Al(oh)3 . vay tong nAl3+ la 0.1+0.04+0.08=0.22 mol =&gt; Cm =2.2 MCau 2: khoi luong kl trong dung dich : 36g &gt; 27.6g kl thu duoc. =&gt; Al het, dung dich chi chua Al 3+ va Fe 2+.ta co nFe2 + = (36-27.6)/56 =0.15 mol.Ap dung dinh luat bao toan dien tich : nAl3+*3=nFe3+*3&nbsp; +&nbsp; nFe3+*1&nbsp; +&nbsp; nCu2+*2&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; hay :&nbsp; 3*m/27 = 0.15*3 + 0.15*1 + 0.3*2 =&gt;m = 10.8 gCau 3: ta co&nbsp; N2O(44)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 8&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; ______&nbsp; &nbsp; &nbsp; 36&nbsp; &nbsp; &nbsp; &nbsp; __&nbsp; =1/1&nbsp; (phuong phap duong cheo)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; N2 (28)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 8=&gt; hon hop gom 2 phan, moi khi chiem 1 phan, nhh=2.688/22.4=0.12 =&gt; N2O=N2=0.06 mol. &nbsp; =&gt;&nbsp; m= 24.84+(0.06*8+0.06*10)*62=91.8g</blockquote>

[QUOTE="quylot1010, post: 30103, member: 24416"] Cau 1: trong lan them dau : OH- ket tua 1 phan Al3+ => nAl3+ =0,1 mol. trong lan them thu 2: OH- lai lam ket tua them mot phan nua =>nAl3+ =0.04 mol Tong oH- them vao : 0.5 mol Tong oH- lam ket tua Al3+ la 0.42 = 3*0.1+3*0.04 => 0.42<0.5 => co 0.08 mol Oh- lam tan Al(oh)3 . vay tong nAl3+ la 0.1+0.04+0.08=0.22 mol => Cm =2.2 M Cau 2: khoi luong kl trong dung dich : 36g > 27.6g kl thu duoc. => Al het, dung dich chi chua Al 3+ va Fe 2+. ta co nFe2 + = (36-27.6)/56 =0.15 mol. Ap dung dinh luat bao toan dien tich : nAl3+*3=nFe3+*3 + nFe3+*1 + nCu2+*2 hay : 3*m/27 = 0.15*3 + 0.15*1 + 0.3*2 =>m = 10.8 g Cau 3: ta co N2O(44) 8 ______ 36 __ =1/1 (phuong phap duong cheo) N2 (28) 8 => hon hop gom 2 phan, moi khi chiem 1 phan, nhh=2.688/22.4=0.12 => N2O=N2=0.06 mol. => m= 24.84+(0.06*8+0.06*10)*62=91.8g [/QUOTE]

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