Answer:
Balance the equation
Fe3O4 + 4H2---> 3Fe +4H2O
Molar mass H2=(2*1)=2g/mol
Mass of H2=54.29grams
mole of H2= 54.29/2=27.15mol
4 mol of H2 = 1 mol Fe3O4
27.15mol of H2= 6.79mol of Fe3O4
Moles of Fe3O4 =6.79mol
Molar mass of Fe3O4= (56×3)+(16×4)
=168+64=232g/mol
Grams of Fe3O4 required to react with 54.29 grams of H2 is 232g/mol× 6.79mol
=1575.28g
