\(n_{NaOH}=\dfrac{8}{40}=0,2mol\); \(n_{Fe}=\dfrac{11,2}{56}=0,2mol\)
2NaOH+H2SO4\(\rightarrow\)Na2SO4+2H2O
Fe+H2SO4\(\rightarrow\)FeSO4+H2
\(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}+n_{Fe}=\dfrac{1}{2}.0,2+0,2=0,3mol\)
m=\(0,3.98=29,4gam\)
\(n_{H_2}=n_{Fe}=0,2mol\rightarrow V_{H_2}=0,2.22,4=4,48l\)