\(Fe+S\rightarrow FeS\)
\(2Al+3S\rightarrow Al_2S_3\)
\(n_S=\frac{12,8}{32}=0,4\left(mol\right)\)
Gọi a là số mol Fe b là số mol Al
\(\left\{{}\begin{matrix}56a+27b=11\\a+1,5b=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
\(\Rightarrow\%n_{Fe}=\frac{0,1}{0,3}.100\%=33,33\%\)