\(n_{Zn}=0,2\left(mol\right)\) \(n_S=0,1\left(mol\right)\)
\(Zn+S-t^0->ZnS\)
Ta có \(\dfrac{0,2}{1}>\dfrac{0,1}{1}\) => Zn dư
\(Zn+2HCl-->ZnCl_2+H_2\)
0,1.......0,2
\(ZnS+2HCl-->ZnCl_2+H_2S\)
0,1..........0,2
\(V_{HCl}=\dfrac{0,2+0,2}{2}=0,2\left(l\right)\)