\(n_{Fe}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\left(1\right)\)
Theo \(pthh\left(1\right):n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
\(n_{H_2SO_4}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=n\cdot22,4=0,1\cdot22,4=2,24\left(l\right)\\ m_{H_2SO_4}=n\cdot M=0,1\cdot98=9,8\left(g\right)\)