2Al + 3S -> Al2S3
0.1 <- 0.15 -> 0.05
Sau pư: Al dư 0.05 mol, S hết, Al2S3: 0,05 mol
2Al + 6HCl -> 2AlCl3 + 3H2
0.05 0.15 0.05 0.15
Al2S3 + 6HCl -> 2AlCl3 + 3H2S
0.05 0.3 0.1 0.15
nH2 = nH2S = 0.15 mol => mỗi khí chiếm 50% thể tích
HCl dư = 0.45 x 0.1 = 0.045 mol
HCl + KOH -> KCl + H2O
0.045 0.045
AlCl3 + 3KOH -> KCl + Al(OH)3
0.15 -> 0.45 0.15
Al(OH)3 + KOH -> KAlO2 + 2H2O
0.005 0.005
=> m Al(OH)3 = (0.15 - 0.005) x 78 = 11.31 g
V HCl bd = (0.45 + 0.045)/ 1 = 0.495 l = 495 ml