\(Fe+2HCl\rightarrow FeCl_2+H_2\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\\ Tacó:\left\{{}\begin{matrix}m_{hh}=56x+27y=8,3\\n_{H_2}=x+\dfrac{3}{2}y=0,25\end{matrix}\right.\\ x=-2,6;y=5,5\)
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