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What mass of Fe3O4 would be needed to

react with 1.38 x 103 mol H2?

Fe3O4 + 4 H2 -> 3 Fe + 4 H2O

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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

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