nHCl= 0,2.1=0,2(mol)
PTHH: 2 M + 6 HCl -> 2 MCl3 + 3 H2 (1)
0,04_________0,12___0,04____0,06(mol)
nNaOH= 0,08.1=0,08(mol)
PTHH: HCl + NaOH -> NaCl + H2O (2)
0,08________0,08_______0,08(mol)
-> nHCl(p.ứ (1))= 0,2- 0,08=0,12(mol)
=> M(M)= mM/nM=1,08/0,04= 27(g/mol)
=> M(III) cần tìm là nhôm (Al=27)
V=V(H2,đktc)=0,06.22,4=1,344(l)