nFe= 11.2/56=0.2 mol
2Fe + 6H2SO4 --> Fe2(SO4)3 + 3SO2 + 6H2O
0.2____0.6__________________0.3
mH2SO4= 0.6*98=58.8g
mddH2SO4= 58.8/200*100%=29.4%
nNaOH=0.2 mol
T=nNaOH/nSO2= 0.2/0.3= 0.66
T\(\le\) 1 => Tạo ra muối NaHSO3
NaOH + SO2 --> NaHSO3
0.2______________0.2
mNaHSO3= 0.2*104=20.8g