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what is the number of moles of Fe2O3 formed when 5.6liter of O2 reacts with 5.6g of Fe?

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2 answers

  1. Balanced Reaction is:
    4 Fe+3O2→2 Fe2O3

    Given,
    5.6liters of O2 participates in the reaction
    1mole → 22.4 lt
    x moles → 5.6lt
    x=​5.6/22.4 = 0.25 moles of O2

    5.6g of Fe participates in the reaction
    1mole → 56g
    x moles → 5.6g
    x=5.6/56 = 0.1 moles of Fe

    In the reaction
    4 Fe+3O2→2 Fe2O3

    4 moles Fe reacts with 3 moles of O2 to give 2 moles of Fe2O3

    As Fe has least mass it acts as a limiting reagent
    So,
    4 moles Fe gives 2 moles of Fe2O3= 2(0.1)/4
    = 0.05 moles

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  2. assuming STP for the O2, you have 5.6/22.4 = 0.25 moles O2

    5.6g = 5.6/55.85 = 0.10 moles Fe

    If your reaction is

    4Fe + 3O2 = 2Fe2O3

    then the reaction consumes elements in the ratio
    (moles O2)/(moles Fe) = 3/4

    Your available reagents are in the ratio of O2/Fe = 0.25/0.10 = 5/2

    So, there is excess O2, and you will wind up with 0.05 moles of Fe2O3, since it takes 2 Fe to produce 1 Fe2O3.

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