\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(FeS+2HCl\rightarrow FeCl_2+H_2S\)
Ta có:
\(\left\{{}\begin{matrix}n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\\n_{FeS}=\frac{17,6}{88}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow V_{khi}=V_{H2}+V_{H2S}=0,1.22,4+0,2.22,4=6,72\left(l\right)\)