nBa(OH)2 = 0.6 mol
nHCl = 2V (mol)
TH1 : Ba(OH)2 dư
Ba(OH)2 + Al + H2O --> Ba(AlO2)2 + 3/2H2
0.12______0.12
=> nBa(OH)2 phản ứng = 0.6 - 0.12 = 0.48 mol
Ba(OH)2 + 2HCl --> BaCl2 + 2H2O
0.48_______0.96
<=> 2V = 0.96
<=> V = 0.48 (l)
TH2: HCl dư
2Al + 6HCl --> 2AlCl3 + 3H2
0.12___0.36
nHCl phản ứng = 2V - 0.36 (mol)
Ba(OH)2 + 2HCl --> BaCl2 + 2H2O
0.6________1.2
<=> 2V - 0.36 = 1.2
<=> V = 0.78 (l)