Em tham khảo nha :
\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{{11,2}}{{22,4}} = 0,5mol\\
hh:Fe(a\,mol),Zn(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,5\\
56a + 65b = 29,8
\end{array} \right.\\
\Rightarrow a = 0,3;b = 0,2\\
{m_{Fe}} = 0,3 \times 56 = 16,8g\\
\% Fe = \dfrac{{16,8}}{{29,8}} \times 100\% = 56,4\% \\
\% Zn = 100 - 56,4 = 43,6\% \\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 1mol\\
{C_{{M_{HCl}}}} = \dfrac{1}{{0,6}} = \dfrac{5}{3}M
\end{array}\)