\(SO_3+H_2O\rightarrow H_2SO_4\\ n_{SO_3}=\dfrac{8}{80}=0,1\left(mol\right)\\ n_{H_2SO_4}=n_{SO_3}=0,1\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{Fe}=n_{H_2}=n_{H_2SO_4}=0,1\left(mol\right)\\ \Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\\ V_{H_2}=0,1.22,4=2,24\left(l\right)\)