PLEASE HELP ME OU How many grams of Fe3O4 are required to react completely with 54.29 grams of H2?

Fe3O4 + H2 → Fe + H2O

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

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