Learn how to find the limiting and excess reactants in a chemical reaction. See example problems that calculate the limiting and excess reactants.
Updated: 02/11/2022
How to Find Limiting and Excess Reactants
Excess Reactant
Consider going grocery shopping for a meal that must be prepared for dinner. There will always be some ingredients left over and some that will be totally consumed once the dish is created, no matter how well planned it is. This situation can be compared to chemical processes that occur all around us. When reactants are allowed to react to generate products, some reactants will be completely consumed while others will be left behind after the reaction is completed. The surplus reactants are referred to as excess reactants since they are in excess of what was necessary. As a result, an excess reactant can be defined as a chemical that is not totally consumed in the reaction.
Another analogy of this can be a worker putting together bicycles in a factory. They must attach two wheels to one bike frame to make a proper bicycle. The process of manufacturing a bicycle can be stated as a chemical equation: {eq}2W + F \rightarrow B_2F {/eq}, where W = wheel, F = frame, and B2F = bicycle.
How many bicycles can be built if there are 20 wheels and 11 frames? Only ten because, despite the fact that there is an 11th frame, there are no wheels remaining to place on it. The concept of an excess reactant refers to this extra frame, which is in surplus.
Limiting Reactant
Pretend you're making chocolate chip cookies. You want to make a full batch of them (24). The recipe makes 24 cookies, but requires two cups of chocolate chips. You have enough of all the other ingredients but only have one cup of chocolate chips. So what ingredient is preventing you from making a full batch? The chocolate chips! You could make a batch with half the chips, but that wouldn't be any good. The lack of chocolate chips is limiting you in making a full batch. It is the limiting factor.
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Limiting and Excess Reactants
In the above example of assembling a bicycle, it can be seen that while one frame was left over, all of the wheels were used up in the process. As a result, the number (or quantity) of wheels determined how many final products (bicycles) could be manufactured, i.e. the wheels set the limit. As a result, limiting reactants are those that limit the amount of other reactants that can combine and the final number of products that can be generated in a chemical process. In a chemical reaction, limiting reactants finish before the other reactants.
How to Find Excess Reactant
Assume a reaction that commences with 10 moles of hydrogen gas (H2) and 7 moles of oxygen gas (O2). The coefficients of a balanced equation for water formation, {eq}2H_2 + O_2 \rightarrow 2H_2O {/eq} , reveal that one mole of H2 and half a mole of O2 are spent for every one mole of water produced. In this example, 5 moles of O2 will react with 10 moles of H2 to produce 5 moles of water. Because there are 7 moles of O2 available when the reaction starts but only 5 are consumed, 2 moles of O2 are left behind. Let's put this into a table to make it easier to understand (Figure 1)
| 2H2 | O2 | 2H2O | |
| At the start of the reaction. | 10 moles | 7 moles | 0 moles |
| Overall Change during the reaction. | - 10 moles | - 5 moles | + 5 moles |
| At the end of the reaction | 0 moles | 2 moles | 5 moles |
Figure 1 : A table showing the change in moles over the course of a chemical reaction.
The third row in the table refers to the quantity consumed (shown through the minus signs) or formed (shown through the plus signs). Because all of the H2 has been used, it is the limiting reactant in this example; however, some O2 remains, making it the excess reactant.
To summarize, Figure 2 depicts the steps to find limiting and excess reactants.
| Step 1 | Write a balanced reaction of the process to establish the coefficients. |
| Step 2 | Equate the reactant coefficients as ratios to find the limiting reactant |
| Step 3 | Identify the reactant that will get consumed completely; this is the limiting reactant. |
| Step 4 | Identify the reactant that will be left; this is the excess reactant. |
Figure 2 : Steps to determining the limiting and excess reactant in a chemical reaction.
How to Find the Limiting Reactant
Figure 2 clearly shows that determining the excess reactant without first determining the limiting one is impossible. This is because only when one of the reactants has been used up can the other be stated to be in excess. The next sections go through specific example problems for determining the limiting reactant.
Identify the Limiting Reactant
When the quantity of the material is given in moles, the method in figure 2 can be used to identify the limiting reactant. The following equation describes a reaction between quartz (silicon dioxide) and hydrogen fluoride.
{eq}SiF_4 (g) + 2 H_2O \rightarrow SiO_2 (s) + 4HF(g) {/eq}
Which of the reactants will be the limiting reactant if 4.5 moles of it are reacted with 6.0 moles of HF? To begin, write the balanced equation for this reaction, which has already been provided in this case.
Step 1: Write the balanced equation.
{eq}SiF_4 (g) + 2 H_2O \rightarrow SiO_2 (s) + 4HF(g) {/eq}
Step 2: Equate the reactant coefficients as ratios to find the limiting reactant.
SiO2(s) : 4HF(g) (this signifies that 4 moles of HF are required to completely consume 1 mole of SiO2)
Step 3: Identify the reactant that will get consumed completely; this is the limiting reagent.
18.0 (4* 4.5) mol of HF must be present for 4.5 mol of SiO2 to completely react. However, one can see in the question that there are only 6.0 moles of HF present, making HF the limiting reactant.
Step 4: Identify the reactant that will be left; this is the excess reagent.
Because HF is the limiting reactant, SiO2 must be the excess reactant.
Limiting Reactant Calculation Steps
The moles of the reactants were provided in the sample problem above. However, when performing such reactions in real life, reactants are usually measured out in grams. In such circumstances, an additional step is required, namely converting grams to moles using the equation (on the list of steps it is shown as step 0). Moles are defined as mass (grams) divided by molar mass (g/mol) i.e.
{eq}Moles = Mass \ (grams)\div Molar \ mass \ (g/mol) {/eq}
If one wants to find the limiting reactant when 150 g of H2 and 1500 g of O2 react to produce electricity in a hydrogen fuel cell, one must first convert the mass from grams to moles and then set up the molar ratios.
Step 0: Convert grams to moles
Moles H2 = (150 g H2) * (1 mol H2/ 2.02 g H2) = 74 moles of H2 (here, 2.02 is the molar mass of H2)
Moles O2 = (1500 g O2) * (1 mol O2/ 32.0 g O2) = 47 moles of O2 (here, 32.0 is the molar mass of O2)
Step 1: Write the balanced equation
{eq}2H_2(g) + O2(g)\rightarrow 2H_2O(g) {/eq}
Step3: Equate the reactant coefficients as ratios to find the limiting reactant
2H2 : O2
The balanced equation's coefficients show that the reaction requires 2 mol of H2 for every 1 mol of O2. As a result, 2 * 47 = 94 mol of H2 are needed to thoroughly react with all of the O2. Because there are only 74 mol of H2, all of the O2 cannot react, leaving H2 as the limiting reactant.
Step 4: Identify the reactant that will be left; this is the excess reagent.
Once H2 has been established as the limiting reactant, it is safe to assume that O2 is the excess reactant.
Excess Reactant Formula
To determine the excess reactant, first determine the limiting reactant, as explained in the previous section. Two more calculations can be performed once the limiting reactant has been identified.
Step 5: Calculate the amount of product that has been generated.
Step 6: Determine how much excess reactant is remaining.
Using the same sample problem as before, the next step is to figure out how much product was generated. Because H2 is the limiting reactant and determines the amount of product generated, this calculation is done in reference to it. A table, similar to that used in figure 1, can be used to visualize the flow of calculations.
| Amount | 2H2 | O2 | 2H2O |
|---|---|---|---|
| At the start of the reaction. | 74 moles | 47 moles | 0 moles |
| Overall Change during the reaction. | - 74 moles | - (74/2) moles | + 74 moles |
| At the end of the reaction | 0 moles | 37 moles | 74 moles |
| In grams (Mass = moles * molar mass) | 0 grams | 37* 32 = 1184 grams | 74 *18 = 1332 grams |
| Excess reactant leftover | - | 1500 - 1184 = 316 grams | - |
Excess Reactant Example
Alternatively, the quantity of the product created and the residual reactants can also be calculated through direct formulas as follows.
Alternate Step 5: Calculate the amount of product that has been generated.
Product in grams = moles of limiting reactant * (mole ratio of product / limiting reactant) * molar mass of product
Grams H2O = (74) * (2/2)* (18.0) = 1332 g
Alternate Step 6: Determine how much excess reactant is remaining
Excess reactant used up in grams = moles of limiting reactant * (mole ratio of excess reactant / limiting reactant) * molar mass of excess reactant
Grams O2 used up = (74)* (1/2)* (32.0) = 1184 g
Grams O2 left = 1500 - 1184 = 316 g
It's worth noting that before calculating the amount of residual surplus, the amount used must first be determined. This amount is then removed from the starting quantity.
Lesson Summary
To summarize, not all reactants are consumed completely in a chemical process. Some reactants are left over and are referred to as excess reactants, while others are completely consumed and are referred to as limiting reactants. The limiting reactants govern the amount of excess reactant that will be consumed and the amount of product that will be generated, when the reaction comes to a halt. It is possible to detect which reactant species is limiting and which is excessive by equating the moles of the reactant species.
To determine how much excess reactant is present and how much product is produced, one must first determine how much limiting reactant is present. This can be done by first converting grams to moles (if necessary) and then equating the coefficients of the chemical process's balanced equation. The molar ratios aid in determining which reactant is limiting and which is excess. The product generated can be determined by equating the moles of the limiting reactant and product or by using a direct formula; Product in grams = moles of limiting reactant * (mole ratio of product / limiting reactant) * molar mass of product. Equating the limiting and excess reactants can also help determine how much of the excess reactant was used up and how much was left.
What Is the Limiting Reactant?
In chemistry, just like real life, there are rarely exact amounts of substances undergoing chemical reactions. There is usually too much of one reactant and not enough of another. The reactant that's used up first and so that no more product can be made is the limiting reactant. The substance that is in excess that doesn't get used up as a reactant is the excess reactant.
If you want to make CO2 and you start with C and O2, the reaction looks like this:
C + O2 --> CO2
One mole of carbon reacts with one mole of O2 to make one mole of CO2. What if I have two moles of C and one mole of CO2? Can I make two moles of CO2? No. Why not? Because I still only have one mole of O2, so I will run out of O2 before I run out of C. O2 is the limiting reactant. This lesson will teach you how to determine the limiting reactant in a reaction and calculate how much excess reactant you have.
Example One
Iron corrodes in the equation 3Fe + 4 H2 O --> Fe3 O4 + 4 H2. If I have 40g of water that react with 150g of iron, what is the limiting reactant? How much excess reactant do I have?
3Fe + 4 H2 O --> Fe3 O4 + 4 H2
You have 40g H2 O, and it reacts with 150g Fe. What is the limiting reactant and what is the mass in grams that is produced? Start with what you're given: 40g H2 O, 150g Fe, ???, Fe3 O4.
A previous lesson taught you how to determine how many grams of substance are in a mole of that substance. For this lesson, I will just tell you:
18g H2 O/1 mole, 55.8g Fe/1 mole, 232g Fe3 O4/1 mole.
What do you need to do first to determine the limiting reactant? You need to change the mass of the reactants to moles of reactants:
- 40g H2 O x (1mole H2 O/18g H2 O) = 2.22 moles H2 O
- 150g Fe x (1 mole Fe/55.8g Fe) = 2.69 moles Fe
Now let's pick one of the reactants to use in our comparison. It doesn't matter which one we pick. Let's use H2 O. We need to compare the moles of H2 O that we have to the mole ratio of Fe to H2 O. You can see that the moles of H2 O will cancel out and this will give us the moles of Fe that we need to use up all of the moles of H2 O that we have.
2.22 moles H2 O x 3 moles Fe/4 moles H2 O = 1.67 moles Fe.
So, for the 2.22 moles of H2 O that we have, we need 1.67 moles of Fe to use it all up in the reaction. Now we could make some nice rules for how to figure out the limiting reactant, but since you probably already have too many rules to learn, let's just think this through. We figured out earlier that we have 2.69 moles Fe available, and we now know that we only need 1.67 moles Fe to use up the 2.22 moles of H2 O that we have. So the H2 O will be used up before the Fe is gone. H2 O is the limiting reactant, the Fe is the excess reactant, and you will have 2.69 - 1.67 = 1.02 moles Fe left over.
To determine how much product Fe3 O4 will be made, multiply the limiting reactant times the mole ratio of product to the limiting reactant and then multiply by the molar mass of the product. Again, you can follow along as the labels cancel out.
2.22 moles H2 O x (1 mole Fe3 O4/4 moles H2 O) x (232g Fe3 O4/1 mole) = 129g Fe3 O4.
Example Two
Zinc and sulfur react to form zinc sulfide. If 100g Zn reacts with 30g S8, what is the limiting reactant? How much product is made? You are given the following information: 8Zn + S8 --> 8ZnS, 65.4g Zn/1 mole Zn, 256.5g S8/1 mole S8, 97.5g ZnS/1 mole ZnS.
Change the masses of the reactants to moles:
- 100g Zn x (1 mole/65.4g Zn) = 1.53 moles Zn
- 30g S8 x (1 mole/256.5g S8) = 0.117 moles S8
Let's use the moles of Zn as our comparison and figure out how much S8 we will need to use up the 1.53 moles of Zn available.
1.53 moles Zn x 1 mole S8/8 moles Zn = 0.191 moles S8. For the 1.53 moles of Zn that we have, we need 0.191 moles of S8. But we only have 0.117 moles of S8 available, so S8 is the limiting reactant as it will run out first. Zn is the excess reactant.
To calculate the amount of Zn left over, we need to find out the amount of Zn used when we have 0.117 moles of S8, and then subtract that amount from the original amount of Zn that we started with.
0.117 moles S8 x (8 moles ZnS/1 mole S8) = 0.936 moles Zn. 1.53 moles Zn available - 0.936 moles Zn used = 0.59 moles Zn in excess. Since S8 is the limiting reactant, we use the limiting reactant amount, 0.1 moles, to find the amount of product produced. We will need the ratio of moles ZnS to moles S8 and the molecular mass of ZnS when determining the amount of product made.
0.117 moles S8 x 8 moles ZnS/1 mole S8) x (97.5g ZnS/1 mole ZnS) = 91.3g ZnS. The total product you can make in this case is 91.3g of zinc sulfide.
Lesson Summary
In a chemical reaction in the real world, there isn't always enough of each reactant in a reaction for every bit of each one to be used up. One of the reactants will be used up before the rest. The reactant that gets used up first is the limiting reactant. The reactant left over when the reaction is complete is the excess reactant.
To determine which of the reactants is the limiting one, choose one, then use stoichiometric ratios to calculate how much you will need of the other one to use up all of the first one you chose. Now, think it through. If you have more than enough of the other reactant, then your first choice is the limiting reactant. If you don't have enough of the second reactant to use up the first one, then the second reactant is the limiting reactant.
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Video Transcript
Limiting Reactant
Pretend you're making chocolate chip cookies. You want to make a full batch of them (24). The recipe makes 24 cookies, but requires two cups of chocolate chips. You have enough of all the other ingredients but only have one cup of chocolate chips. So what ingredient is preventing you from making a full batch? The chocolate chips! You could make a batch with half the chips, but that wouldn't be any good. The lack of chocolate chips is limiting you in making a full batch. It is the limiting factor.
What Is the Limiting Reactant?
In chemistry, just like real life, there are rarely exact amounts of substances undergoing chemical reactions. There is usually too much of one reactant and not enough of another. The reactant that's used up first and so that no more product can be made is the limiting reactant. The substance that is in excess that doesn't get used up as a reactant is the excess reactant.
If you want to make CO2 and you start with C and O2, the reaction looks like this:
C + O2 --> CO2
One mole of carbon reacts with one mole of O2 to make one mole of CO2. What if I have two moles of C and one mole of CO2? Can I make two moles of CO2? No. Why not? Because I still only have one mole of O2, so I will run out of O2 before I run out of C. O2 is the limiting reactant. This lesson will teach you how to determine the limiting reactant in a reaction and calculate how much excess reactant you have.
Example One
Iron corrodes in the equation 3Fe + 4 H2 O --> Fe3 O4 + 4 H2. If I have 40g of water that react with 150g of iron, what is the limiting reactant? How much excess reactant do I have?
3Fe + 4 H2 O --> Fe3 O4 + 4 H2
You have 40g H2 O, and it reacts with 150g Fe. What is the limiting reactant and what is the mass in grams that is produced? Start with what you're given: 40g H2 O, 150g Fe, ???, Fe3 O4.
A previous lesson taught you how to determine how many grams of substance are in a mole of that substance. For this lesson, I will just tell you:
18g H2 O/1 mole, 55.8g Fe/1 mole, 232g Fe3 O4/1 mole.
What do you need to do first to determine the limiting reactant? You need to change the mass of the reactants to moles of reactants:
- 40g H2 O x (1mole H2 O/18g H2 O) = 2.22 moles H2 O
- 150g Fe x (1 mole Fe/55.8g Fe) = 2.69 moles Fe
Now let's pick one of the reactants to use in our comparison. It doesn't matter which one we pick. Let's use H2 O. We need to compare the moles of H2 O that we have to the mole ratio of Fe to H2 O. You can see that the moles of H2 O will cancel out and this will give us the moles of Fe that we need to use up all of the moles of H2 O that we have.
2.22 moles H2 O x 3 moles Fe/4 moles H2 O = 1.67 moles Fe.
So, for the 2.22 moles of H2 O that we have, we need 1.67 moles of Fe to use it all up in the reaction. Now we could make some nice rules for how to figure out the limiting reactant, but since you probably already have too many rules to learn, let's just think this through. We figured out earlier that we have 2.69 moles Fe available, and we now know that we only need 1.67 moles Fe to use up the 2.22 moles of H2 O that we have. So the H2 O will be used up before the Fe is gone. H2 O is the limiting reactant, the Fe is the excess reactant, and you will have 2.69 - 1.67 = 1.02 moles Fe left over.
To determine how much product Fe3 O4 will be made, multiply the limiting reactant times the mole ratio of product to the limiting reactant and then multiply by the molar mass of the product. Again, you can follow along as the labels cancel out.
2.22 moles H2 O x (1 mole Fe3 O4/4 moles H2 O) x (232g Fe3 O4/1 mole) = 129g Fe3 O4.
Example Two
Zinc and sulfur react to form zinc sulfide. If 100g Zn reacts with 30g S8, what is the limiting reactant? How much product is made? You are given the following information: 8Zn + S8 --> 8ZnS, 65.4g Zn/1 mole Zn, 256.5g S8/1 mole S8, 97.5g ZnS/1 mole ZnS.
Change the masses of the reactants to moles:
- 100g Zn x (1 mole/65.4g Zn) = 1.53 moles Zn
- 30g S8 x (1 mole/256.5g S8) = 0.117 moles S8
Let's use the moles of Zn as our comparison and figure out how much S8 we will need to use up the 1.53 moles of Zn available.
1.53 moles Zn x 1 mole S8/8 moles Zn = 0.191 moles S8. For the 1.53 moles of Zn that we have, we need 0.191 moles of S8. But we only have 0.117 moles of S8 available, so S8 is the limiting reactant as it will run out first. Zn is the excess reactant.
To calculate the amount of Zn left over, we need to find out the amount of Zn used when we have 0.117 moles of S8, and then subtract that amount from the original amount of Zn that we started with.
0.117 moles S8 x (8 moles ZnS/1 mole S8) = 0.936 moles Zn. 1.53 moles Zn available - 0.936 moles Zn used = 0.59 moles Zn in excess. Since S8 is the limiting reactant, we use the limiting reactant amount, 0.1 moles, to find the amount of product produced. We will need the ratio of moles ZnS to moles S8 and the molecular mass of ZnS when determining the amount of product made.
0.117 moles S8 x 8 moles ZnS/1 mole S8) x (97.5g ZnS/1 mole ZnS) = 91.3g ZnS. The total product you can make in this case is 91.3g of zinc sulfide.
Lesson Summary
In a chemical reaction in the real world, there isn't always enough of each reactant in a reaction for every bit of each one to be used up. One of the reactants will be used up before the rest. The reactant that gets used up first is the limiting reactant. The reactant left over when the reaction is complete is the excess reactant.
To determine which of the reactants is the limiting one, choose one, then use stoichiometric ratios to calculate how much you will need of the other one to use up all of the first one you chose. Now, think it through. If you have more than enough of the other reactant, then your first choice is the limiting reactant. If you don't have enough of the second reactant to use up the first one, then the second reactant is the limiting reactant.
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Frequently Asked Questions
What is the purpose of an excess reactant?
The purpose of the excess reactant is to ensure that the limiting reactant gets fully used up during the course of the chemical reaction.
How do you find the excess reactant?
To find the the excess reactant, one must first determine the limiting reactant. This is done by equating the coefficients of the reactants. The ratio setup hints to the reactant that will deplete first, making it the limiting reactant, and the other one is the one in excess.
How do you calculate a LR?
To find the limiting reactant (LR), begin by balancing the chemical equation and converting all quantities to moles. After that, compare the mole ratios of the reactants. This comparison will reveal which substance is present in lower quantities than necessary and is the the LR.
What is an excess and limiting reagent?
In a chemical reaction, the chemical that is consumed fully is the limiting reagent whereas the the chemical that is in surplus is the excess reagent.
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BackHow to Find Limiting & Excess Reactants | Limiting & Excess Reactant Formula, Calculation Steps & Example
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