Hỗn hợp A$\left\{ \begin{align}
& {{H}_{2}}:1mol \\
& C{{O}_{2}}:amol \\
\end{align} \right.$
$\begin{align}
& Fe+2HCl\to FeC{{l}_{2}}+{{H}_{2}} \\
& 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \\
& MgC{{O}_{3}}+2HCl\to MgC{{l}_{2}}+C{{O}_{2}}+{{H}_{2}}O \\
& a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a \\
\end{align}$
$\begin{align}
& 2Fe+6{{H}_{2}}S{{O}_{4}}\to F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+3S{{O}_{2}}+6{{H}_{2}}O \\
& 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1,5 \\
& MgC{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\to MgS{{O}_{4}}+C{{O}_{2}}+{{H}_{2}}O \\
& a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a \\
\end{align}$
Ta có: $\frac{2+44a}{1+a}:\frac{64\times 1,5+44a}{1,5+a}$ = $\frac{1}{3,6875}$ $\Rightarrow a=0,5$
%$Fe=57,14$%
%$MgC{{O}_{3}}$ = 42,86%