Use the balanced equation from number four above to determine the limiting reactant if we had a mixture of 5.0 moles of Fe and 4.0 moles of O2.

2Fe + 3O₂ → 2Fe₂O₃

Limiting reactant: O₂

2Fe + O₂ → 2FeO

Limiting reactant: Fe

Explanation:

2Fe + 3O₂ → 2Fe₂O₃

2Fe + O₂ → 2FeO

These are the possible reactions:

In first case, 2 moles of Fe need 3 mol of oyxgen to react

If I have 5 moles of Fe, I will need (5.3) / 2 = 7.5 moles of O₂

Then, the oxygen is my limiting (I only have 4 moles)

3 moles of O₂ need 2 moles of Fe

If I have 4 moles of O₂, I will need (4.2) / 3 = 2.66 moles (I have 5)

Fe, is the reactant in excess.

For second case, 2 moles of Fe need 1 mol of O₂, to react.

If I have 5 moles of Fe, I will need (5.1) / 2 = 2.5 moles of O₂

I have 4 moles of oxygen, so now it is my excess.

1 mol of O₂ need 2 moles of Fe, to react

If I have 4 moles of O₂, I will need the double of amount, 8 moles of Fe.

I have 5 moles, then the Fe is my limtiing reactant.