Em tham khảo nha :
\(\begin{array}{l}
2Al + 3CuS{O_4} \to A{l_2}{(S{O_4})_3} + 3Cu\\
Fe + CuS{O_4} \to FeCuS{O_4} + Cu\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
hh:Al(a\,mol),Fe(b\,mol),Cu(c\,mol)\\
{n_{Cu}} = \dfrac{{35,2}}{{64}} = 0,55mol\\
\Rightarrow \dfrac{3}{2}a + b + c = 0,55(1)\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4mol\\
\Rightarrow \dfrac{3}{2}a + b = 0,4(2)\\
\text{Lấy (1) trừ (2) }\Rightarrow c = 0,15mol\\
{m_{Cu}} = 0,15 \times 64 = 9,6g
\end{array}\)
