FexOy+yH2→xFe+yH2O
Gọi mH2O=a
mH2SO4=100.98/100=98g
mdd sau pứ=a+100
C%H2SO4 sau pứ=98-3,405=94,595%
→98/(a+100)=94,595/100
→9800=94,595a+9459,5
→94,595a=340,5
→a=3,6
nH2O=3,6/18=0,2mol
Fe+H2SO4→FeSO4+H2
nH2=3,36/22,4=0,15mol
→nFe=0,15mol
x:y=nFe:nH2O=0,15:0,2=3:4
→Fe3O4