Question

the number of moles of Fe2O3 formed when 5.6lit 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

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.