Bảo toàn khối lượng:
$\begin{align}
& {{m}_{KCl{{O}_{3}}}}={{m}_{r}}+{{m}_{{{O}_{2}}}}\Rightarrow {{m}_{{{O}_{2}}}}=98-93,2=4,8g \\
& {{n}_{{{O}_{2}}}}=4,8:32=0,15mol \\
\end{align}$
${{m}_{{{O}_{2}}}}+{{m}_{KL}}={{m}_{Y}}\Rightarrow {{m}_{KL}}=15,6-4,8=10,8g$
$\left\{ \begin{align}
& Mg:a \\
& Fe:b \\
\end{align} \right.(mol)$
$24a+56b=10,8$
Bảo toàn e:
$\begin{align}
& Mg\to M{{g}^{2+}}+2e \\
& a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2a \\
& Fe\to F{{e}^{3+}}+3e \\
& b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3b \\
\end{align}$
$\begin{align}
& {{O}_{2}}+4e\to 2{{O}^{2-}} \\
& 0,15\,\,\,\,\,0,6 \\
& {{S}^{+6}}+2e\to {{S}^{+4}} \\
& \,\,\,\,\,\,\,\,\,0,05\leftarrow 0,025 \\
\end{align}$
Vậy:
2a + 3b = 0,6 + 0,05 =0,65
Giải hpt được a = 0,1; b = 0,15
% Mg = $\frac{24\times 0,1}{10,8}\times 100$% = 22,22%