\(n_{H_2}=\dfrac{6,048}{22,4}=0,27\left(mol\right)\\ Đặt:a=n_{Fe};b=n_{Al}\left(a,b>0\right)\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow hpt:\left\{{}\begin{matrix}56a+27b=9,42\\a+1,5b=0,27\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,12\\b=0,1\end{matrix}\right.\\ \Rightarrow\%m_{Fe}=\dfrac{0,12.56}{9,42}.100\approx71,338\%\)