$n_{H_2} = 5,6:22,4 = 0,25$ (mol)
Dat $n_{Al} = a$ (mol), $n_{Fe} = b$ (mol)
PTPU
$$2Al + 6 HCl -> 2AlCl_3 + 3H_2$$
$$ Fe + 2 HCl -> FeCl_2 + H_2$$
Theo de bai ta co
$$\begin{cases}
27a + 56b = 8,3\\
1,5a + b = 0,25
\end{cases}$$
Vay $a = b = 0,1$.
$m_{Al} = 27.0,1 = 2,7$ (g), $m_{Fe} = 56.0,1 = 5,6$ (g)