nH2 = 5.6/22.4 = 0.25 (mol)
Fe + H2SO4 => FeSO4 + H2
...........0.25.........................0.25
VH2SO4 = 0.25/1 = 0.25 (l)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4}=n_{H_2}=0,25\left(mol\right)\)
\(\Rightarrow V=\dfrac{n}{C_M}=\dfrac{0,25}{1}=0,25\left(l\right)\)