What mass of Fe3O4 would be needed to
react with 1.38 x 103 mol H2?
Fe3O4 + 4 H2 -> 3 Fe + 4 H2O
1.38*10^3 I assume
one mol Fe3O4 for every 4 mol H2
1380/4 = 345 mols of Ferric oxide
345 * (3*56+4*16) grams
react with 1.38 x 103 mol H2?
Fe3O4 + 4 H2 -> 3 Fe + 4 H2O
one mol Fe3O4 for every 4 mol H2
1380/4 = 345 mols of Ferric oxide
345 * (3*56+4*16) grams