Identify the limiting reactant and excess reactant

Question

Identify the limiting reactant and excess reactant when 78.20 g of K reacts with 141.8 g of Cl₂ to produce KCl. How many grams of the excess reactant is left over?

                 2 K + Cl₂ --> 2 KCl 

Answer 1

This is a multi-step problem that can be somewhat lengthy, so I will discuss it in parts.

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". If we are lacking enough of one reactant to make product, then the reaction can't proceed, and that reactant will be called the limiting reactant as it limits how much product we can make.

So, for this example, we have 2K + Cl₂ -> 2KCl. When we read this equation, it tells us that 2 moles of K react with 1 mole of Cl₂ to form 2 moles of KCl.

Our reactants are 2K and Cl₂ and our product is 2KCl

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 grams of each reactant we have. So, we must convert our grams of reactants to moles of reactants so that we can compare them to our balanced equation.

1. 78.20 g of K = ? moles of K
2. The molar mass of K is 39.0983 g. So, 1 mole of K = 39.0983 g of K.
3. If we have 78.20 g, we can convert it to moles by dividing the grams we have of K by the number of grams in 1 mole of K.
4. 78.20 g/39.0983 g = 2.00 moles of K
5. 141.8 g of Cl₂ = ? moles of Cl₂
6. The molar mass of Cl₂ is 70.906 g. So, 1 mole of Cl₂ = 70.906 g of Cl₂.
7. If we have 141.8 g of Cl₂, we can convert it to moles by dividing the grams we have of Cl₂ by the number of grams in 1 mole of Cl₂.
8. 141.8 g/70.906 g = 1.9998 moles of Cl

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 K
2. Based on our balanced equation, 2 moles of K will yield 2 moles of KCl.
3. In our previous step, we determined that the grams we had of K = 2 moles of K. So, if we fully react all of our K, we should yield 2 moles of KCl.
4. Reacting Cl
5. Based on our balanced equation, 1 mole of Cl will yield 2 moles of KCl.
6. In our previous step, we determined that the grams we had of Cl = 1.9998 moles of Cl. So, if we fully react all 1.9998 moles of our Cl at a 1:2 ratio, we should make 3.9996 moles of KCl.

However, we know from determining our theoretical yield of K, that if we were to fully react all 78.20 g of K (or 2 moles of K), we could only yield a max of 2 moles of KCl based on our balanced equation. Even though we have enough Cl₂ to make more, we are limited by how much K we have as it will be used up first.

So, K is our limiting reagent.

Step 4 - If K is our limiting reagent, and we know that we have excess Cl₂, we can determine how much excess Cl₂ we have.

We know that we have 141.8 g of Cl₂ or 1.9998 moles of Cl₂. Cl₂ reacts in a 1:2 ratio where 1 mole of Cl₂ yields 2 moles of KCl. Even though we have enough Cl₂ to make 3.9996 moles of KCl, we only have enough K to make 2 moles of KCl. So, we need to know how much Cl₂ we will use to make 2 moles of KCl.

Since we know that Cl₂ reacts in a 1:2 ratio where 1 mole of Cl₂ yields 2 moles of KCl, then 2 moles of KCl will need 1 mole of Cl₂. We have 1.9998 moles. So in this reaction, we will only use 1 of those 1.9998 moles leaving us with 0.9998 moles of Cl₂.

We can determine how many grams 0.9998 moles of Cl₂ is by converting it back to grams using the molar mass. In this case, we would multiply the molar mass of Cl₂ (70.906 g) by the number of moles we will have leftover (0.9998 moles).

So, 0.9998 moles of Cl₂ multiplied by 70.906 g = 70.89 g of Cl.

Therefore, our limiting reactant is K, and our excess reactant is Cl₂. We will have 70.89 g of Cl₂ remaining.

*Note, in this problem, I did not round to avoid any error. However, you can round as it wouldn't change the answer much.