Em tham khảo nha :

\(\begin{array}{l}
1)\\
a)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,15mol\\
{m_{Mg}} = n \times M = 0,15 \times 24 = 3,6g\\
{m_{Cu}} = 10 - 3,6 = 6,4\\
2)\\
a)\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,2mol\\
{m_{Mg}} = n \times M = 0,2 \times 65 = 13g\\
{m_{Cu}} = 16,2 - 13 = 3,2g\\
3)\\
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{2,8}}{{22,4}} = 0,125mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,125mol\\
{m_{Fe}} = n \times M = 0,125 \times 56 = 7g\\
{m_{Cu}} = 10 - 7 = 3g\\
\% Fe = \dfrac{7}{{10}} \times 100\%  = 70\% \\
\% Cu = 100 - 70 = 30\% \\
4)\\
a)\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
\text{Dung dịch sau phản ứng có màu xanh lam}\\
b)\\
{n_{HCl}} = V \times {C_M} = 0,2 \times 1 = 0,2mol\\
{n_{CuO}} = \dfrac{{{n_{HCl}}}}{2} = 0,1mol\\
{m_{CuO}} = n \times M = 0,1 \times 80 = 8g\\
\% CuO = \dfrac{8}{{20,8}} \times 100\%  = 38,46\% \\
\% Cu = 100 - 38,46 = 61,54\% 
\end{array}\)