Em tham khảo nha :
\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
hh:Fe(a\,mol),Zn(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,2\\
56a + 65b = 12,1
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,1\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
\% Fe = \dfrac{{5,6}}{{12,1}} \times 100\% = 46,3\% \\
\% Zn = 100 - 46,3 = 53,7\% \\
b)\\
{n_{FeS{O_4}}} = {n_{Fe}} = 0,1mol\\
{m_{FeS{O_4}}} = 0,1 \times 152 = 15,2g\\
{n_{ZnS{O_4}}} = {n_{Zn}} = 0,1mol\\
{m_{ZnS{O_4}}} = 0,1 \times 161 = 16,1g\\
{m_m} = 15,2 + 16,1 = 31,3g
\end{array}\)