Sửa đề : 13.9 (g)
\(n_{Al}=a\left(mol\right),n_{Fe}=b\left(mol\right)\)
\(\Rightarrow m=27a+56b=13.9\left(1\right)\)
\(n_{H_2}=\dfrac{7.84}{22.4}=0.35\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{H_2}=1.5a+b=0.35\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.2\)
\(\%Al=\dfrac{0.1\cdot27}{13.9}\cdot100\%=19.42\%\)
\(\%Fe=100-19.42=80.58\%\)