Câu 1.
nHNO[SUB]3[/SUB] = 0,5 mol
nN[SUB]2[/SUB]O = 0,045 mol => nH[SUP]+[/SUP] pư = 0,45 mol
=> nH[SUP]+[/SUP] tạo NH[SUB]4[/SUB][SUP]+[/SUP] = 0,05 mol => nNH[SUB]4[/SUB][SUP]+[/SUP] = 0,005 mol
=> m = 12 + 62(0,36 + 0,04) + 0,005*80 = 37,2 gam
Câu 2.
Br[SUB]2[/SUB] + 2NaI --> 2NaBr + I[SUB]2[/SUB]
=> nBr[SUB]2[/SUB] = (a – b)/94 => nNaI = (a – b)/47
Cl[SUB]2[/SUB] + 2NaBr --> 2NaCl + Br[SUB]2[/SUB]
=> nCl[SUB]2[/SUB] = (b – c)/89 => nNaBr = 2(b – c)/89
=> nNaBr trong hỗn hợp ban đầu = 2(b – c)/89 – (a – b)/47 = 2(a – b)/89 – (a – b)/47 = 5(a – b)/4183
=> 150(a – b)/47 + 103[5(a – b)/4183] = a
=> 13350(a – b) + 515(a – b) = 4183a =>
13865a – 13865b = 4183a
9682a = 13865b => a : b = 13865 : 9682
=> %(m) NaBr = 515*100/13865 = 3,71%
Câu 3.
Gọi a, b là nồng độ của dung dịch X và dung dịch Y
=> nBa(OH)[SUB]2[/SUB] = 3aV[SUB]1[/SUB] (mol) ; nAl[SUB]2[/SUB](SO[SUB]4[/SUB])[SUB]3[/SUB] = bV[SUB]1[/SUB] (mol)
Al[SUB]2[/SUB](SO[SUB]4[/SUB])[SUB]3[/SUB] + 3Ba(OH)[SUB]2[/SUB] --> 2Al(OH)[SUB]3[/SUB] + 3BaSO[SUB]4[/SUB]
Kết tủa max => nBa(OH)[SUB]2[/SUB] = 3nAl[SUB]2[/SUB](SO[SUB]4[/SUB])[SUB]3[/SUB] => 3aV[SUB]1[/SUB] = 3bV[SUB]1[/SUB] => a = b
=> m = 156aV[SUB]1[/SUB] + 699aV[SUB]1[/SUB] = 855aV[SUB]1[/SUB] => 0,9m = 769,5aV[SUB]1[/SUB]
nBa(OH)[SUB]2[/SUB] = aV[SUB]2[/SUB] ; nAl[SUB]2[/SUB](SO[SUB]4[/SUB])[SUB]3[/SUB] = bV[SUB]1[/SUB] = aV[SUB]1[/SUB]
Al[SUB]2[/SUB](SO[SUB]4[/SUB])[SUB]3[/SUB] + 3Ba(OH)[SUB]2[/SUB] --> 2Al(OH)[SUB]3[/SUB] + 3BaSO[SUB]4[/SUB]
+ Giả sử Al[SUB]2[/SUB](SO[SUB]4[/SUB])[SUB]3[/SUB] dư => nAl(OH)[SUB]3[/SUB] = 2aV[SUB]2[/SUB]/3 (mol) ; nBaSO[SUB]4[/SUB] = aV[SUB]2[/SUB]
=> 769,5aV[SUB]1[/SUB] = 52aV[SUB]2[/SUB] + 233aV[SUB]2[/SUB] => V[SUB]2[/SUB] : V[SUB]1[/SUB] = 0,27
+ Giả sử có hòa tan kết tủa.
2Al(OH)[SUB]3[/SUB] + Ba(OH)[SUB]2[/SUB] --> Ba(AlO[SUB]2[/SUB])[SUB]2[/SUB] + 4H[SUB]2[/SUB]O
nBaSO[SUB]4[/SUB] = 3aV[SUB]1[/SUB] (mol)
=> nAl(OH)[SUB]3[/SUB] thu được = 2aV[SUB]1[/SUB] – 2(aV[SUB]2[/SUB] – 3aV[SUB]1[/SUB]) = 8aV[SUB]1[/SUB] – 2aV[SUB]2[/SUB]
=> 769,5aV[SUB]1[/SUB] = 699aV[SUB]1[/SUB] + 78(8aV[SUB]1[/SUB] – 2aV[SUB]2[/SUB]) => V[SUB]2[/SUB] : V[SUB]1[/SUB] = 3,55
