Em tham khảo nha :
\(\begin{array}{l}
a)\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
{n_{S{O_2}}} = \dfrac{{22,4}}{{22,4}} = 1mol\\
hh:Fe(a\,mol),Cu(b\,mol)\\
\left\{ \begin{array}{l}
56a + 64b = 48\\
\dfrac{3}{2}a + b = 1
\end{array} \right.\\
\Rightarrow a = 0,4;b = 0,4\\
{m_{Fe}} = 0,4 \times 56 = 22,4g\\
\% Fe = \dfrac{{22,4}}{{48}} \times 100\% = 46,7\% \\
\% Cu = 100 - 46,7 = 53,3\% \\
b)\\
{n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{Fe}}}}{2} = 0,2mol\\
{n_{CuS{O_4}}} = {n_{Cu}} = 0,4mol\\
{m_m} = 0,2 \times 400 + 0,4 \times 160 = 144g
\end{array}\)
