The Limiting Reactant Concept

In the previous lesson, we learned how to use the weight and mole relationships in a chemical reaction to predict how much of a chemical could be produced (or would be needed) given some amount of one of the other chemicals. In each of those problems, however, we assumed that we had enough of the other chemicals so that the reaction would proceed until all the reactants were used up. What if one of the reactant chemicals were to run out first? The reaction would have to stop and we might have some of the other reactant left over. The chemical that runs out first would limit how much of the product(s) we could make; that is the concept behind limiting reactant problems. The reactant chemical that is used up first is called the limiting reactant.

Example 1

To be able to answer problems involving limiting reactants, first we need to have a way to determine the known ratio between the reactants. Sometimes you'll have a chemical equation to work with (we'll get to that in the next section of this lesson). Sometimes you'll simply be told how much of each chemical is known to react.

To start, let's consider the reaction of hydrogen and oxygen to make water. Thinking back to prior work in the lab, you observed that these elements react in a 2:1 ratio by volume. We can express this ratio with the following word equation and table. We will ignore the volume of water for now and focus on the reactants.

hydrogen + oxygen → water

2 mL + 1 mL ---

What does this equation mean? Obviously at one level, it means that 2 mL of hydrogen react fully with 1 mL of oxygen. More generally, it means that any time these two elements react, a complete reaction will require exactly twice as much hydrogen as oxygen by volume. As long as this is the case, there is no need to think about limiting reactants.

But we can easily imagine mixing hydrogen and oxygen in other proportions. There is absolutely nothing preventing you as a chemist from mixing any volume of hydrogen with any other volume of oxygen. And figuring out what will happen in that case requires applying the concept of the limiting reactant.

Let's start with a somewhat silly idea. Instead of mixing 2 mL of hydrogen with 1 mL of oxygen, let's imagine mixing 2 mL of hydrogen with ... 0 mL of oxygen. That is, don't add any oxygen at all. Which of our reactants would run out first?

Obviously the answer is "oxygen," because the question is a bit absurd. There isn't any oxygen to "run out." But, in a certain sense, we would call oxygen our "limiting reactant" here, in that the lack of oxygen limits how long the reaction can run.

To make it somewhat less silly, this time, let's imagine mixing our 2 mL of hydrogen with a tiny volume of oxygen ... say, 0.1 mL. Once again, we ask the question at the heart of the limiting reactant concept: "which reactant will run out first."

The answer is, once again, "oxygen," so oxygen is still the limiting reactant. One way to think about this is that hydrogen and oxygen react in a 2:1 ratio, so 0.1 mL of oxygen will use up 0.2 mL of hydrogen. Since we have 2 mL of hydrogen available, what will happen is that the oxygen will run out, and at that point there will be 1.8 mL of hydrogen left over. This makes hydrogen the excess reactant. Once the oxygen runs out, the reaction stops, and no more water can be produced. Later, we will do problems using stoichiometry to calculate the amount of product produced in a limiting reactant setting, but for now we are just focusing on the reactants.

Now, imagine we mix 2 mL of hydrogen with 1.5 mL of oxygen, and once again ask "if these amounts of these substances react, which will run out first?" We can approach this problem a couple different ways:

• First approach: 1.5 mL of oxygen will require 3 mL of hydrogen to react fully, because of the 2:1 ratio of their reaction. Because we only have 2 mL of hydrogen, it will run out before the oxygen is used up.

• Second approach: 2 mL of hydrogen would use up 1 mL of oxygen, again because of the ratio involved. Since we have 1.5 mL of oxygen, there is oxygen to spare, meaning the hydrogen will be used up first

Both these approaches lead to the same answer: hydrogen is now the limiting reactant. It will run out first, at which point the reaction will cease. Both approaches are correct, and this illustrates a feature of the limiting reactant concept: there are almost always numerous ways to approach a given problem.

Note: in our final example, hydrogen is the limiting reactant even though there is more of it present. This is because it is used up twice as fast in the reaction.

Example 2

Let's return to another reaction that should be familiar from Lesson 3. The word equation and mass relationships for this reaction are shown below. Note: once again, all we have is a word equation and mass data. Later we will see how to use a balanced equation to determine limiting reactants. You need one or the other.

methane + chlorine → methyl chloride + hydrogen chloride

16.0 g + 61.0 g → 40.5 g + 36.5 g

In this reaction, the ratio of chlorine required to fully react with methane is 61.0 grams to 16.0 grams. If I told you that I were to mix 61.0 grams of chlorine with 22.0 g of methane, you could probably tell me without doing any calculations that the chlorine would be the limiting reactant. This is because I know instantly that I have more methane than is needed to fully react with the chlorine. You could maybe even tell me that at the end of the reaction I would have 6.0 g of methane left over (because 61.0 g of chlorine uses up 16.0 g of methane, which we can subtract from 22.0 g).

However, what if I were to combine 75 g of methane with 300 g of chlorine? Which reactant would be limiting now? It's not obvious at all. We will need a more precise way of examining the question.

For any amounts of methane and chlorine that we mix (or other reactants in another reaction), we can determine which of the two substances is limiting by picking one of them and asking two questions: 1) how much of the other would we need to react with all of the one we picked? and 2) do we have that much of the other available?

Let’s pick the methane and ask, first, “how much chlorine would we need to react with all 75 g of the methane?” The question we can answer using one of the methods you learned in lesson 3. (I'll show you the conversion factor approach, but you could use a proportion instead.) Chlorine reacts with methane in the ratio 61 g to 16 g.

The numbers that go into the ratio, the 61.0 and 16.0, come from the information given in the problem. It will be important to keep straight the two sets of numbers – those that reflect the proper ratios of reactants and products to one another, and the set of arbitrary numbers you are being asked about. One of the most common mistakes that students make when they are doing stoichiometry problems is to mix up those sets of numbers. Remember that the known ratio is what goes in the conversion faction, not the new "given" numbers.

Multiplying that known ratio times 75 g of methane tells us that we would need about 285 g of chlorine to react with all 75 g methane. The second question is, “do we have that much chlorine?” Since we have 300 g and only 285 g are needed, we do have enough chlorine to react with all 150 g of methane. Therefore, we will run out of methane first. Methane is the limiting reagent and therefore the chlorine is the excess reagent.

We would have come to the same conclusion by asking, “do we have enough methane to react with all 300 g of chlorine?” The calculation shows that 78.7 g methane would be needed, slightly more than the 75 g we have. Again, we would discover that the methane is limiting.

We can also calculate the mass of chlorine that would be left over at the end of the reaction. To do this, we simply subtract the chlorine consumed from the chlorine present, as shown below. The answer is 15 g of chlorine.

300 g - 285 g = 15 g

Finally, we can also calculate the amount of product (we'll do methyl chloride, though we could also calculate for hydrogen chloride) that would be made. The idea here is this: we have identified the limiting reactant as methane, so it is the reactant that will run out first. At that point, production of methyl chloride ceases. So we can do stoichiometry starting with our 75 g of methane to calculate the amount of methyl chloride produced at the point the reaction ends. The process is one you have done many times in Lesson 3.

As you are hopefully coming to see, while the limiting reactant concept can be tricky, and the problems a bit chaotic, there is nothing very new here in terms of the mathematical techniques to be used. Limiting reactant problems are just stoichiometry problems, with the added layer that there are multiple reactants involved. In the next section, we will see a few techniques for approaching them, especially when working with balanced equations. But you have already seen examples of all the major objectives for this Lesson:

• Determining a limiting reactant given amounts of two reactants

• Determining the amount of product produced given amounts of two reactants

• Determining the amount of excess reactant consumed and/or left over in such a situation