nNaOH=3/40=0.075mol
nFe=11.2/56=0.2mol
2NaOH + H2SO4 -> Na2SO4 + 2H2O
(mol)0.075 0.0375
mH2SO4 pứ = 0.0375*98=3.675g
Fe + H2SO4 -> FeSO4 + H2
(mol) 0.2 0.2 0.2
mH2SO4 dư = 0.2*98=19.6g
a) mH2SO4 bđ = m =19.6+3.675=23.275g
b) VH2=0.2*22.4=4.48l