$n_{H_2}=\frac{11,2}{22,4}=0,5(mol)$
$2Al+6HCl→2AlCl_3+3H_2↑$
$Fe+2HCl→FeCl_2+H_2↑$
$Mg+2HCl→MgCl_2+H_2↑$
⇒$n_{HCl}=2.n_{H_2}=2.0,5=1(mol)$
Theo định luật bảo toàn khối lượng:
$⇒m_{hh}+m_{HCl}=m_{muối}+m_{H_2}$
$⇔m_{hh}+1.36,5=53+0,5.2$
$⇔m_{hh}=17,5(g)$