Em tham khảo nha :
\(\begin{array}{l}
a)\\
Fe + HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,25mol\\
{m_{Fe}} = 0,25 \times 56 = 14g\\
{m_{Cu}} = 21 - 14 = 7g\\
b)\\
\% Fe = \dfrac{{14}}{{21}} \times 100\% = 66,67\% \\
\% Cu = 100 - 66,67 = 33,33\% \\
c)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,5mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,5}}{{0,2}} = 2,5M\\
d)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
{n_{Cu}} = \dfrac{7}{{64}} = 0,109375mol\\
{n_{S{O_2}}} = {n_{Cu}} = 0,109375 mol\\
{V_{S{O_2}}} =0,109375 \times 22,4 = 2,45l
\end{array}\)