Đáp án:

 50,72% và 49,28%

Giải thích các bước giải:

\(\begin{array}{l}
2Mg + {O_2} \to 2MgO\\
3Fe + 2{O_2} \to F{e_3}{O_4}\\
n{O_2} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
hh:Mg(a\,mol),Fe(b\,mol)\\
24a + 56b = 9,2\\
0,5a + \frac{{2b}}{3} = 0,15\\
 \Rightarrow a = \dfrac{{17}}{{90}};b = \dfrac{1}{{12}}\\
\% mFe = \dfrac{{\dfrac{1}{{12}} \times 56}}{{9,2}} \times 100\%  = 50,72\% \\
\% mMg = 100 - 50,72 = 49,28\% 
\end{array}\)

$n_{HCl}=3,36/22,4=0,15mol$

$PTHH :$

$3Fe+2O_2\overset{t^o}\to Fe_3O_4$

$2Mg+O_2\overset{t^o}\to 2MgO$

Gọi $n_{Fe}=a;n_{Mg}=b$

$\text{Ta có :}$

$m_{hh}=56a+24b=9,2g$

$n_{O_2}=\dfrac{2}{3}.a+0,5b=0,15mol$

$\text{Ta có hpt :}$

$\left\{\begin{matrix}
56a+24b=9,2 & \\ 
a+b=0,15 & 
\end{matrix}\right.$

$⇔\left\{\begin{matrix}
a=\dfrac{1}{12} & \\ 
b=\dfrac{17}{90} & 
\end{matrix}\right.$

$⇒\%m_{Fe}=\dfrac{\dfrac{1}{12}.56.100\%}{9,2}=50,72\%$

$\%m_{Mg}=100\%-50,72\%=49,28\%$