Em tham khảo nha :
\(\begin{array}{l}
a)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
{n_{S{O_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
hh:Cu(a\,mol);Fe(b\,mol)\\
\left\{ \begin{array}{l}
64a + 56b = 12\\
a + \frac{3}{2}b = 0,25
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,1\\
{m_{Cu}} = 64 \times 0,1 = 6,4g\\
\% Cu = \dfrac{{6,4}}{{12}} \times 100\% = 53,3\% \\
\% Fe = 100 - 53,3 = 46,7\% \\
b)\\
{n_{NaOH}} = 0,5 \times 2 = 1mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{1}{{0,25}} = 4\\
\Rightarrow\text{Phản ứng tạo ra 1 muối } N{a_2}S{O_3}\\
2NaOH + S{O_2} \to N{a_2}S{O_3} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,25mol\\
{m_{N{a_2}S{O_3}}} = 0,25 \times 126 = 31,5g\\
c)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
{n_{{H_2}}} = {n_{Fe}} = 0,1mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
d)\\
{n_{{H_2}S{O_4}}} = 2{n_{Cu}} + 3{n_{Fe}} = 0,5mol\\
{m_{{H_2}S{O_4}}} = 0,5 \times 98 = 49g\\
{m_{dd{H_2}S{O_4}}} = \dfrac{{49 \times 100}}{{98}} = 50g
\end{array}\)