Em tham khảo nha :
\(\begin{array}{l}
a)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{S{O_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{Cu}} = {n_{S{O_2}}} = 0,1mol\\
{m_{Cu}} = 0,1 \times 64 = 6,4g\\
\% Cu = \dfrac{{6,4}}{{9,1}} \times 100\% = 70,33\% \\
\% Al = 100 - 70,33 = 29,67\% \\
b)\\
{n_{NaOH}} = 0,1 \times 1 = 0,1mol\\
T = \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,1}}{{0,1}} = 1\\
\Rightarrow\text{Phản ứng tạo ra 1 muối } NaHS{O_3}\\
NaOH + S{O_2} \to NaHS{O_3}\\
{n_{NaHS{O_3}}} = {n_{NaOH}} = 0,1mol\\
{m_{NaHS{O_3}}} = 0,1 \times 104 = 10,4g
\end{array}\)