$n_{HCl}=3,36/22,4=0,15mol$
$PTHH :$
$3Fe+2O_2\overset{t^o}\to Fe_3O_4$
$2Mg+O_2\overset{t^o}\to 2MgO$
Gọi $n_{Fe}=a;n_{Mg}=b$
$\text{Ta có :}$
$m_{hh}=56a+24b=9,2g$
$n_{O_2}=\dfrac{2}{3}.a+0,5b=0,15mol$
$\text{Ta có hpt :}$
$\left\{\begin{matrix}
56a+24b=9,2 & \\
a+b=0,15 &
\end{matrix}\right.$
$⇔\left\{\begin{matrix}
a=\dfrac{1}{12} & \\
b=\dfrac{17}{90} &
\end{matrix}\right.$
$⇒\%m_{Fe}=\dfrac{\dfrac{1}{12}.56.100\%}{9,2}=50,72\%$
$\%m_{Mg}=100\%-50,72\%=49,28\%$