How to Calculate the Limiting Reactant of a Chemical Reaction

In a chemical reaction, the limiting reactant (RL) is the reactant that is in the smallest stoichiometric proportion . What this means is that it corresponds to the reactant that runs out first as the reaction progresses. When this happens, the reaction cannot continue, so the amount of other reactants that can be consumed is limited, as well as the amount of products that can be formed, hence its name.

Why is it important to determine the limiting reactant?

In view of the fact that the limiting reagent is the one that determines, when finished, the amounts of all the other substances that can effectively participate in the reaction, it is then the most important from the point of view of stoichiometric calculations. In fact, all stoichiometric calculations must be carried out based solely on the limiting reactant, or some other quantity that has been calculated based on it, since doing it with any of the other reactants (which are called reactants in excess), will lead to an excess calculation error.

As an example, let’s consider a recipe for a cake that calls for:

• 1 cup of milk
• 2 cups flour
• 1 cup of sugar, and
• 4 eggs.

Now suppose that in the refrigerator we have

• 5 cups of milk
• 8 cups of flour
• 2 cups of sugar, and
• 20 eggs.

How many cakes can we make with these ingredients?

This type of problem is very similar to that of a chemical reaction for which we have a recipe (given by the balanced or balanced chemical equation), we can have variable amounts of ingredients (which become the reactants), and one or more products.

If we analyze separately how many cakes we can prepare with each of the ingredients we have, we will obtain different possible amounts of cakes:

• Since each cake requires only 1 cup of milk, with 5 cups of milk we could make 5 cakes.
• The 8 cups of flour are enough to prepare 4 cakes.
• Each cake has 2 cups of sugar, so with 2 cups we can make only 2 cakes.
• With 20 eggs we could make 5 cakes, since each one requires 4 eggs.

It is evident that the maximum number of cakes that we can prepare in this case is 2, since we do not have enough sugar to prepare 4, let alone 5 cakes. That is, after we finish preparing the second cake, we will run out of sugar, so we will not be able to continue preparing more cakes, even if we have plenty of the other ingredients.

In this case, sugar represents the “limiting ingredient” in our cake factory. The concept of the limiting reagent, as well as the way to identify it, is exactly the same. With that said, let’s see how the limiting reactant is calculated or determined in a chemical reaction.

When should we determine which is the limiting reactant and when not?

Before learning how to determine what the limiting reactant is, we must know in which situations it is necessary to do so. In principle, all stoichiometric calculations must be carried out starting with the limiting reagent. However, in some situations it is not necessary to determine it either because it is already known in advance what it is, or because, with the information available, there is no other solution than to assume what the limiting reactant is.

The rules for whether or not we should determine the limiting reactant before starting the stoichiometric calculations are:

• If there is only one reactant, there is no concept of a limiting reactant, so determining it is not necessary.
• If we react a reactant in the presence of an excess of another (because the statement of a problem explicitly indicates so, for example), then the first will be the limiting reactant and it is not necessary to determine it.
• In the event that we want to calculate how much product can be obtained from a given amount of a single reactant, regardless of whether other reactants are involved in the reaction, we carry out the calculations assuming that the first reactant is the limiting reactant and that we have enough of all the other reagents involved.
• On the other hand, if a chemical reaction involves two or more reactants and we have fixed or limited amounts of two or more of them, we must always determine which is the limiting reactant before carrying out the other calculations .

Methods to determine the limiting reagent of a chemical reaction

The limiting reactant is a concept that scares many basic chemistry students, but it doesn’t have to be. Problems involving the limiting reagent are easy to recognize and can all be solved in the same way. It’s just about finding a quick and easy way to determine what the limiting reactant is, and then use it in all the stoichiometric calculations we need to carry out.

Below are three different ways to determine the limiting reactant. Some are more intuitive and are similar to the cake example. Others are less intuitive, but are more practical and easier to use, especially in complex reactions involving many reactants. The idea is that, by the end of this article, you will have learned how to determine the limiting reactant in any situation, and that you have chosen one of the three methods for everyday use in all the stoichiometric calculations you will need to perform in the future.

The explanation of the three methods is based on the same problem that is stated below and that involves three reagents of which we have certain or limited quantities.

Limiting Reactant Calculation Problem

Given the reaction for the formation of potassium phosphate:

Determine the amount of this compound that could be formed if 19.55g of potassium, 3.10g of phosphorus, and 32.0g of oxygen gas are reacted. Data: the relative atomic masses of the elements involved are: K:39.1; P:31.0 and 0:16.0.

Method 1: The method how much do I have? – how much I need?

Since we have limited amounts of all three reactants, we must determine which is the limiting reactant before carrying out stoichiometric calculations to obtain the amount of potassium phosphate. The first method we’ll look at is to determine how much of each reactant is needed to fully consume the other reactants, and then compare this result to how much of the reactant we actually have.

If when carrying out the calculation it turns out that we have more than what we need, then that will be the excess reagent. On the other hand, if we have less than we need to react with the other reactants, then that will be the limiting reactant since it is not enough.

NOTE: It should be noted that this method only allows you to compare two reagents at a time to determine the limiting factor between them. In cases like the present example, which involve more than two reagents, the comparison must be carried out consecutively until determining which is the global limiting reagent. It should also be noted that the calculations can be carried out in terms of masses or moles. In this case, it will be carried out in mass, and in the next two methods the calculations will be carried out in moles.

The method how much do I have? – how much I need? It consists of the following steps:

Step 1: Determine the molar masses of all reactants involved

In the present case, the molar masses are:

                MMK ₌ 39.1 g/mol

                MM _P =31.0 g/mol

                MM _O2 = 2×16.0 g/mol = 32.0 g/mol

Step 2: Determine the masses of all reactants, if not available.

In this case, we already know the masses of all reactants. These are:

                mK = _19.55g

                m _P = 3.10g

                mO2 = _32.0g

Step 3: Select two of the reagents involved

In this case, we will start with potassium (K) and phosphorous (P), but the order in which the reactants are chosen is not important.

Step 4: Calculate the amount of the first that would react with the given amount of the second.

At this point we will carry out the first stoichiometric calculation. These are calculations of the hypothetical amounts that would be needed of each reagent in order to fully consume the other. That is, we will determine, first of all, how much potassium we would need to completely consume the 3.10 g of phosphorus that we have. This calculation is carried out by means of a simple stoichiometric relationship:

This result means that we need 11.73 g of potassium to completely consume the 3.10 g of phosphorus that we have.

Step 5: Calculate the amount of the second that would react with the given amount of the first.

This step is the opposite of the previous step. That is, we will calculate the amount of phosphorus that we would need to completely consume all the potassium that we have.

This result means that we need 5.17 g of phosphorus to completely consume the 19.55 g of potassium that we have.

Step 6: Fill in a Have/Need table and choose the limiting and excess reagent

This table contains the two reactants we are comparing, the actual amounts of each that we have, and the amounts needed that we just determined in steps 4 and 5. Additionally, some people add a column with the difference between what we have and what we need, since the sign of this difference can be used to quickly determine what the RL is, although it is preferable to determine it logically to avoid errors.

As we can see, in the case of potassium we have more than we need to completely consume phosphorus, which is why potassium is an excess reactant. This automatically implies that, between these two reagents, phosphorus is the limiting reagent. This can also be deduced by analyzing the results for phosphorus. To consume all the potassium, we would need 5.17 g of phosphorus, but we only have 3.10 g. This means that the phosphorus we have is not enough to consume all the potassium, so it runs out first, ie, it is the limiting reactant between the two.

Another easy way to determine the limiting reagent almost without thinking is by selecting the one whose difference T – N is negative.

At this point we call phosphorus the partial limiting reactant since we don’t yet know if it will still be the limiting reactant once we compare it to oxygen. That’s what the next step is about.

Step 7: Repeat steps 4, 5 and 6 with the previous limiting reagent and another reagent.

Since we determined that phosphorus is the RL between it and potassium, we must now compare it to all the other reactants involved in the reaction. In this case, this involves comparing it to oxygen. To do this, we repeat steps 4, 5 and 6 but using P and O ₂ .

Since there are no more reagents left that we have not compared, we conclude that the overall limiting reagent (or, simply, the limiting reagent) is phosphorus .

Method 2: Calculation of a product

This method is based on the same principle as the pie example we saw earlier. It consists, simply, in determining the amount of the same product that can be obtained from the given amount of each reactant. In the end, the limiting reactant is the one that produces the least amount of that product. Stoichiometric calculations can be carried out in masses or in moles. The only thing that changes is the use of molar masses in the stoichiometric ratios that are used in the calculations. Since the previous method was carried out using masses, we will implement this method using moles, but it should be remembered that it can also be applied to masses.

The steps are the following:

Step 1: Determine all the molar masses of the reactants.

This is the same first step as the previous method so we won’t repeat it here.

Step 2: Determine the moles of all reactants, if not available.

This calculation consists of dividing the masses by the respective molar masses:

                nK ₌ 19.55g / 39.1g/mol = 0.500 mol

                nP ₌ 3.10g / 31.0g/mol = 0.100 mol

                nO2 = 32.0g / 32.0g/mol = 1.00 _mol

Step 3: Calculate the moles of the same product that can be produced with each reactant.

Using the stoichiometric ratios in moles, which are obtained directly from the balanced chemical equation, we calculate the hypothetical moles that we could obtain from each reactant if it were completely consumed:

Step 4: The limiting reactant will be the one that produces the least amount of product.

We can summarize the calculations we have made in the following table:

As expected, the limiting reagent turned out to be phosphorus again.

Method 3: Method of stoichiometric proportions

This method consists of determining the stoichiometric proportion in which each reactant is found in relation to the adjusted chemical equation. Then, by definition, the limiting reactant is the one with the smallest proportion. This ratio is determined by dividing the number of moles of each reactant by its stoichiometric coefficient.

Of all, this is the easiest method to use, as it can be done very quickly and without much thought. The first two steps are the same as those of the previous method, and all that remains is to add the calculation of the stoichiometric ratio:

Once again, the limiting reagent turns out to be phosphorus.

Final comments

The steps for the determination of the limiting reagent presented here must be adapted in the case of reactions in aqueous solutions in which concentrations and volumes of solution are used instead of masses or moles. The same can be said of the case in which one works with gases and one has the pressure or the volume of a gas. In any case, the only thing that would change would be the process of calculating the moles or mass, but everything else would remain the same.

References

Bolívar, G. (2019, June 8). Limiting and Excess Reactant: How It’s Calculated and Examples . lifer. https://www.lifeder.com/reactivo-limitante-en-exceso/

Chang, R. (2021). Chemistry (11th ^ed .). MCGRAW HILL EDUCATION.

Limiting Reagent Examples . (nd). Químicas.net. https://www.quimicas.net/2015/10/ejemplos-de-reactivo-limitante.html

The yields of the reactions. (2020, October 30). https://espanol.libretexts.org/@go/page/1822