\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PT :
2Mg + O2 \(\xrightarrow[]{t^o}\) 2MgO
a 0,5a
4Fe + 3O2 \(\xrightarrow[]{t^o}\) 2Fe2O3
b 0,75b
Ta có hpt :
\(\left\{{}\begin{matrix}24a+56b=9,2\\0,5a+0,75b=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,15\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\%m_{Mg}=\dfrac{0,15.24}{9,2}.100\%\approx39,13\%\)
\(\Rightarrow\%m_{Fe}=100\%-39,13\%=60,87\%\)
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