PTHH: 2M + 6HCl -> 2$MCl_{3}$ + 3$H_{2}$ (1)
HCl + NaOH -> NaCl + $H_{2}$$O$ (2)
Từ PTHH (2) => $n_{HCl(2)}$ = $n_{NaOH}$ = 0,08 (mol)
=> $n_{HCl(1)}$ = $n_{HCl }$ - $n_{HCl(2)}$ = 0,12 (mol)
Từ PTHH (1) => $n_{M}$ = $\frac{1}{3}$ $n_{HCl(1)}$ = 0,04 (mol)
=> $M_{M}$ = $\frac{1,08}{0,04}$ = 27 (g/mol)
=> M là Al
Từ PTHH (1) => $n_{H2}$ = $\frac{1}{2}$ = 0,02 (mol)
=> $V_{H2}$ = 0,02 * 22,4 = 0,448(l)