Em tham khảo nha :
\(\begin{array}{l}
a)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
Mg + 2{H_2}S{O_4} \to MgS{O_4} + S{O_2} + 2{H_2}O\\
{n_{S{O_2}}} = \dfrac{{7,84}}{{22,4}} = 0,35mol\\
hh:Mg(a\,mol);Cu(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,35\\
24a + 64b = 12,4
\end{array} \right.\\
\Rightarrow a = 0,25;b = 0,1\\
{m_{Mg}} = 0,25 \times 24 = 6g\\
\% Mg = \dfrac{6}{{12,4}} \times 100\% = 48,4\% \\
\% Cu = 100 - 48,4 = 51,6\% \\
b)\\
{n_{{H_2}S{O_4}}} = 2{n_{Cu}} + 2{n_{Mg}} = 0,7mol\\
{C_{{M_{{H_2}S{O_4}}}}} = \dfrac{{0,7}}{{0,2}} = 3,5M\\
c)\\
{n_{MgS{O_4}}} = {n_{Mg}} = 0,25mol\\
{m_{MgS{O_4}}} = 0,25 \times 120 = 30g\\
{n_{CuS{O_4}}} = {n_{Cu}} = 0,1mol\\
{m_{CuS{O_4}}} = 0,1 \times 160 = 16g\\
{m_m} = 30 + 16 = 46g
\end{array}\)