Em tham khảo nha :
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{0,7437}}{{24,79}} = 0,03mol\\
hh:Fe(a\,mol),Mg(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,03\\
56a + 24b = 1,04
\end{array} \right.\\
\Rightarrow a = 0,01;b = 0,02\\
{m_{Fe}} = 0,01 \times 56 = 0,56g\\
{m_{Mg}} = 1,04 - 0,56 = 0,48g\\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,06mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,06}}{{0,2}} = 0,3M
\end{array}\)
