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Trả lời:
Đặt $n_{Al}=a;\,\,n_{Fe}=b$
$n_{O_2}=\dfrac{5,6}{22,4}=0,25\,(mol)$
$Al^0\rightarrow Al^{+3}+3e\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O_2^0+4e\rightarrow2O^{-2}\\\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,25\,\,\,\,\,\,\,1\\Mg^0\rightarrow Mg^{+2}+2e\\\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2b$
$n_{eTĐ}=1\,(mol)$
$⇒3a+2b=1$
Khối lượng của hỗn hợp là $10,2\,(g)$
$⇒27a+24b=10,2$
Ta có hệ: $\begin{cases}3a+2b=1\\27a+24b=10,2\end{cases}$
$⇔\begin{cases}a=0,2\\b=0,2\end{cases}⇔\begin{cases}m_{Al}=0,2.27=5,4\,(g)\\m_{Mg}=0,2.24=4,8\,(g)\end{cases}.$