Question

The number of moles of Fe2O3 formed when 5.6 liters of O2 reacts with 5.6g of Fe

Answer 1

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.

Answer 2

Answer:

0.05

Explanation:

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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