Identify the limiting reactant and excess reactant

Question

Identify the limiting reactant and excess reactant when 18 moles of Fe react with 18 moles of O₂ to produce Fe₃O₄. How many moles of the excess reactant are left over?

                 3 Fe + 2 O₂ -->  Fe₃O₄

Answer 1

First, a limiting reactant is defined as being the reactant in a chemical reaction that will be used up first. In a chemical reaction, we are only able to make as much "product" as we have "reactants".

So, for this example, we have 3Fe + 2O₂ -> Fe₃O₄. When we read this equation, it tells us that 3 moles of Fe react with 2 moles of O₂ to form 1 mole of Fe₃O₄.

Our reactants are 2Fe and 2O₂ and our product is Fe₃O₄

Step 1 - Always check to make sure your equation is balanced. In this case, it is. So we are good to go.

Step 2 - Determine how many moles of each reactant we have. In the problem, we were told how many moles of each reactant we have. So, we just need to write those down.

1. 18 moles of Fe
2. 18 moles of O₂

Step 3 - Determine how much product we can make if we were to fully react all of each reactant we have. This is what we call theoretical yield as, in theory, it is the amount of our product that we should be able to make if we were able to fully react all of our reactants.

1. Reacting Fe
2. Based on our balanced equation, 3 moles of Fe will yield 1 mole of Fe₃O₄. So, if we fully react all of our Fe (18 moles), we should yield 6 moles of Fe₃O₄.
3. Reacting O2
4. Based on our balanced equation, 2 moles of O₂ will yield 1 mole of Fe₃O₄. So, if we fully react all 18 moles of our O₂ at a 2:1 ratio, we should make 9 moles of Fe₃O₄.

However, we know from determining our theoretical yield of Fe, that if we were to fully react all 18 moles g of Fe, we could only yield a max of 6 moles of Fe₃O₄ based on our balanced equation. Even though we have enough O₂ to make more, we are limited by how much Fe we have as it will be used up first.

So, Fe is our limiting reagent.

Step 4 - If Fe is our limiting reagent, and we know that we have excess O₂, we can determine how much excess O₂ we have.

We know that we have 18 moles of O₂. O₂ reacts in a 2:1 ratio where 2 moles of O₂ yields 1 mole of Fe₃O₄. Even though we have enough O₂ to make 9 moles of Fe3O4, we only have enough Fe to make 6 moles of Fe3O4. So, we need to know how much O2 we will use to make 6 moles of Fe3O4.

Since we know that O₂ reacts in a 2:1 ratio where 2 moles of O2 yields 1 mole of Fe3O4, then 6 moles of Fe3O4 will need 12 moles of O2. We have 18 moles. So in this reaction, we will only use 12 of those 18 moles leaving us with 6 moles of O2.

We can determine how many grams 6 moles of O2 is by converting it back to grams using the molar mass. In this case, we would multiply the molar mass of O2 (32 g) by the number of moles we will have leftover (6 moles).

So, 6 moles of O2 multiplied by 32 g = 192 g of O2.

Therefore, our limiting reactant is Fe, and our excess reactant is O2. We will have 192 g of O2 remaining or 6 moles of O2.