Em tham khảo nha :
\(\begin{array}{l}
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
{n_{S{O_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
hh:Cu(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
64a + 56b = 12\\
a + \dfrac{3}{2}b = 0,25
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,1\\
{m_{Cu}} = 0,1 \times 64 = 6,4g\\
\% Cu = \dfrac{{6,4}}{{12}} \times 100\% = 53,3\% \\
\% Fe = 100 - 53,3 = 46,7\%
\end{array}\)
