\(n_{FeS}=\dfrac{8,8}{88}=0,1\left(mol\right)\\ n_{HCl}=0,1.1=0,1\left(mol\right)\)
PTHH: FeS + 2HCl ---> FeCl2 + H2S
LTL: \(0,1>\dfrac{0,1}{2}\) => FeS dư
\(n_{H_2S}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\\ V_{H_2S}=0,05.22,4=1,12\left(l\right)\)