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Trả lời:
Đặt $n_{Al}=a;\,\,n_{Fe}=b$
$n_{H_2}=\dfrac{5,6}{22,4}=0,25\,(mol)$
$Al^0\rightarrow Al^{+3}+3e\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2H^{+1}+2e\rightarrow H_2^0\\\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,5\,\,\,\,\,\,\,\,0,25\\Fe^0\rightarrow Fe^{+2}+2e\\\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2b$
$n_{eTĐ}=0,5\,(mol)$
$⇒3a+2b=0,5$
Khối lượng của hỗn hợp là $8,3\,(g)$
$⇒27a+56b=8,3$
Ta có hệ: $\begin{cases}3a+2b=0,5\\27a+56b=8,3\end{cases}⇔\begin{cases}a=0,1\\b=0,1\end{cases}$
$⇔\begin{cases}m_{Al}=0,1.27=2,7\,(g)\\m_{Fe}=0,1.56=5,6\,(g)\end{cases}⇔\begin{cases}\%m_{Al}=\dfrac{2,7.100\%}{8,3}=32,53\%\\\%m_{Fe}=\dfrac{5,6.100\%}{8,3}=67,47\%\end{cases}$
