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

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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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