a) \(Fe+S--->FeS\)
x--------x(mol)
\(Zn+S-->ZnS\)
y-----y(mol)
b) \(n_S=\frac{9,6}{32}=0,3\left(mol\right)\)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}56x+65y=18,6\\x+y=0,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)
\(\%m_{Fe}=\frac{0,2.56}{18,6}.100\%=60,22\%\)
\(\%m_{Zn}=100-60,22=39,78\%\)
Tỉ lệ %
\(\%Fe:\%Zn=60,22:39,78\)