Lesson 4: Limiting Reagents

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 reagent problems. The reactant chemical that is used up first is called the limiting reagent.

In this section, we'll look at some examples and then there are a few more in the workbook for you to try on your own (the answers will be at the end of this page).

First Example | Second Example | Practice

First Example: Limiting Reagents

To be able to answer problems involving limiting reagents, 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.

Let's look at Exercise 1 in your workbook. We are told that hydrogen reacts with oxygen in a 2:1 ratio by volume to make water; so for every 2 mL of hydrogen that react, 1 mL of oxygen will react. That is our known ratio. If we are given other amounts of hydrogen and oxygen, we can compare them to the known ratio and determine which one will be used up first.

The new amount of hydrogen matches the amount of hydrogen in the known ratio - 2 mL. We know that 2 mL of hydrogen reacts with 1 mL of oxygen. So out of the 10 mL of oxygen in the new amount, 1 mL will be used up and there will be 9 mL left over. The 2 mL of hydrogen is our limiting reagent and the oxygen is our "excess reagent."

This is similar to the previous one we looked at. 2 mL of hydrogen will only react with 1 mL of oxygen; we have 4 mL of oxygen, so 1 mL will be used up and 3 mL of oxygen will be left over. Again, the 2 mL of hydrogen is our limiting reagent and the oxygen is our excess reagent.

Now we have a new amount of hydrogen to work with; 4 mL is just double the amount from the known ratio. That means we would need twice the known amount of oxygen, or 2 mL of oxygen. Since that is the amount of oxygen we have, there won't be anything left over. Both of our reactants are limiting and there won't be any excess.

Let's look at one more together.

Starting again with the 4 mL of hydrogen, we know from the previous example that we'd need 2 mL of oxygen. This time, however, we only have 1 mL of oxygen; that's not enough to use up all of the hydrogen. The 1 mL of oxygen is our limiting reagent; it will react with 2 mL of hydrogen, so we will have 2 mL of hydrogen left over.

Try your hand with the rest of the examples in Exercise 1; the answers are in the workbook. If you have difficulty, get help before moving on.

Top of page

Second Example: Limiting Reagents

From the given information, we know that 16 g of methane react with 61 g of chlorine. What would happen if we mixed 18 g of methane with 61 g of chlorine? The reaction would continue until we ran out of chlorine. And since we started with more methane than we needed, there would be some left over: 2 grams.

Since the reaction continues until we run out of chlorine, the chlorine is what limits the amount of product that forms, and so is called the “limiting reagent.” We have extra methane left over, so the methane is called the ”excess reagent.”

Notice that the limiting reagent is not necessarily the one that there is less of. There is more than three times as much as chlorine as methane, and yet it is the chlorine that is limiting. The limiting reagent is the one we have less of than we need to react with the other entire reagent present. The concept is quite general. We might have three or four or more reactants in a given reaction. The limiting reagent will always be the one we run out of first.

For any amounts of methane and chlorine that we mix, 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?

 

In other words, we might ask, “do we have enough methane to react with all 600 g of chlorine?” We would need to determine first, how much methane we would need to react with 600 g of chlorine, and then compare that amount with the 150 g of methane we have. Alternately, we also might figure out how much chlorine we would need to react with 150 g of methane and compare that number to the 600 g of chlorine we have.

Let’s pick the methane and ask, first, “how much chlorine would we need to react with all 150 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.

150 g methane	x	61 g chlorine	=	572 g chlorine
		16 g methane

The numbers that go into the ratio, shown in red here, 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 150 g of methane tells us that we would need about 572 g of chlorine to react with all 150 g methane. The second question is, “do we have that much chlorine?” Since we have 600 g and only 572 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. (However, had there been a third reactant, this result would tell us only that the chlorine is not limiting. We would need to do a similar calculation to determine which of the remaining two reactants was limiting.)

We would have come to the same conclusion by asking, “do we have enough methane to react with all 600 g of chlorine?”

The calculation shows that we would need about 157 g of methane, which is 7 g more than we have. Since we don't have enough methane to use up all 600 g of chlorine, we find that methane is the limiting reagent.

You can sometimes determine the limiting reagent without doing any calculations, as we did with the hydrogen and oxygen examples. For example, suppose we had started with equal amounts of methane and chlorine, or with less chlorine than methane. In either case, we know without doing any calculations that we need more chlorine than methane but don’t have it, so the chlorine must be limiting.

Since the limiting reagent is the one that determines when the reactions stops, it is the limiting reagent that determines how much of any given product will form.

150 g methane	x	40.5 g methyl chloride	=	308 g methyl chloride
		16 g methane

When we allow 150 g of methane to react with 600 g of chlorine, we use the 150 g of methane to determine that about 308 g of methyl chloride will be produced.

As the limiting reagent limits the amount of each of the products that form, it also limits the amount of the other reactants that are used up. In any stoichiometry problem, therefore, it is the mass of the limiting reagent that must be used to determine the amounts of any of the other reactants and products that are sought.

Top of page

Practice

Limiting reagent problems are rarely identified explicitly as such, that is, few such problems include the words “limiting reagent.” When you gain some sophistication in solving stoichiometry problems, you will have formed the habit of considering the question, “what is the limiting reagent?” In the mean time, the best indication that you must actively consider which reactant is limiting is when amounts of two different reactants are given in the same problem. When a problem states the amount of only one of the reactants, you must assume the reactant is limiting.

Whenever a problem tells you the amounts of two of the reactants, you must first figure out which of them is the limiting reagent, and then use that reactant to determine the mass of any product that will form or of any other reactant that will be consumed. Exercises 2 and 3 show you some examples and give you some practice problems to try. Exercise 2 shows you a couple more options for techniques to use to find the limiting reagent; after you look them over, pick the one that makes the most sense to you and then practice using it consistently. Exercise 3 has a few problems for you to work out on your own. You will find the answers below.

Answers to Ex. 3:

a. 27 g of magnesium chloride (chlorine is limiting)
b. no, 110 g of sulfur would be needed
c. 28 g of carbon dioxide, (oxygen is limiting)

If you weren't able to solve these problems, get help before moving on. In the next section, we'll use balanced equations as the starting point for solving limiting reagent problems.

Top of page