Em tham khảo nha :
\(\begin{array}{l}
a)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{m_{Cu}} = 0,64g\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
hh:Mg(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,1\\
24a + 56b = 4
\end{array} \right.\\
\Rightarrow a = 0,05;b = 0,05\\
{m_{Mg}} = 0,05 \times 24 = 1,2g\\
\% Cu = \dfrac{{0,64}}{{4,64}} \times 100\% = 13,8\% \\
\% Mg = \dfrac{{1,2}}{{4,64}} \times 100\% = 25,9\% \\
\% Fe = 100 - 25,9 - 13,8 = 60,3\% \\
b)\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,1mol\\
{m_{{H_2}S{O_4}}} = 0,1 \times 98 = 9,8g\\
{m_{{\rm{dd}}{{\rm{H}}_2}S{O_4}}} = \frac{{9,8 \times 100}}{{24,5}} = 40g
\end{array}\)