3Fe + 2O2 → Fe3O4 You are given 9 moles of Fe and 4 moles of O2. Because you are given amounts of both reactants, this is a limiting reactant problem: You are given 9:4 of Fe to O2 in moles and what the stoichiometry of the reaction needs is 3:2 . Clearly, you have excess Fe and O2 is the LR.
Amount of O2 determines product formed and Fe used in reaction:
4 moles O2 (1 Fe3O4/2O2) = 2 moles Fe3O4
4 moles O2 (3 Fe/2O2) = 6 moles Fe used 9-6=3 moles Fe left
No O2 left.
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