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### Course: AP®︎/College Chemistry>Unit 4

Lesson 4: Stoichiometry

# Limiting reactant and reaction yields

## Limiting reactant and theoretical yield

It’s a classic conundrum: We have five hot dogs and four hot dog buns. How many complete hot dogs can we make?
Assuming that hot dogs and buns combine in a $1:1$ ratio, we can make four complete hot dogs. Once we run out of buns, we'll have to stop making complete hot dogs. In other words, the hot dog buns limit the number of complete hot dogs we can produce.
In much the same way, a reactant in a chemical reaction can limit the amounts of products formed by the reaction. When this happens, we refer to the reactant as the limiting reactant (or limiting reagent). The amount of a product that is formed when the limiting reactant is fully consumed in a reaction is known as the theoretical yield. In the case of our hot dog example, we already determined the theoretical yield (four complete hot dogs) based on the number of hot dogs buns we were working with.
Enough about hot dogs, though! In the next example, we'll see how to identify the limiting reactant and calculate the theoretical yield for an actual chemical reaction.

## Example 1: Using the limiting reactant to calculate theoretical yield

A sample of $\mathrm{Al}\left(s\right)$ reacts with a sample of $\mathrm{Cl}{\phantom{A}}_{2}\left(g\right)$ according to the equation shown below.
$2\phantom{\rule{0.167em}{0ex}}\mathrm{Al}\left(s\right)+3\phantom{\rule{0.167em}{0ex}}\mathrm{Cl}{\phantom{A}}_{2}\left(g\right)\to 2\phantom{\rule{0.167em}{0ex}}\mathrm{AlCl}{\phantom{A}}_{3}\left(s\right)$
What is the theoretical yield of $\mathrm{AlCl}{\phantom{A}}_{3}$ in this reaction?
To solve this problem, we first need to determine which reactant, $\mathrm{Al}$ or $\mathrm{Cl}{\phantom{A}}_{2}$, is limiting. We can do so by converting both reactant masses to moles and then using one or more mole ratios from the balanced equation to identify the limiting reactant. From there, we can use the amount of the limiting reactant to calculate the theoretical yield of $\mathrm{AlCl}{\phantom{A}}_{3}$.

### Step 1: Convert reactant masses to moles

Let's start by converting the masses of $\mathrm{Al}$ and $\mathrm{Cl}{\phantom{A}}_{2}$ to moles using their molar masses:
$\begin{array}{rl}2.80\phantom{\rule{0.278em}{0ex}}\overline{)\text{g Al}}& ×\frac{1\phantom{\rule{0.278em}{0ex}}\text{mol Al}}{26.98\phantom{\rule{0.278em}{0ex}}\overline{)\text{g Al}}}=1.04×{10}^{-1}\phantom{\rule{0.278em}{0ex}}\text{mol Al}\\ \\ 4.15\phantom{\rule{0.278em}{0ex}}\overline{){\text{g Cl}}_{2}}& ×\frac{1\phantom{\rule{0.278em}{0ex}}{\text{mol Cl}}_{2}}{70.90\phantom{\rule{0.278em}{0ex}}\overline{){\text{g Cl}}_{2}}}=5.85×{10}^{-2}\phantom{\rule{0.278em}{0ex}}{\text{mol Cl}}_{2}\end{array}$

### Step 2: Find the limiting reactant

Now that we know the quantities of $\mathrm{Al}$ and $\mathrm{Cl}{\phantom{A}}_{2}$ in moles, we can determine which reactant is limiting. As you'll see below, there are multiple ways to do so, each of which uses the concept of the mole ratio. All of the methods give the same answer, though, so you can choose whichever approach you prefer!
Method 1: For the first method, we'll determine the limiting reactant by comparing the mole ratio between $\mathrm{Al}$ and $\mathrm{Cl}{\phantom{A}}_{2}$ in the balanced equation to the mole ratio actually present. In this case, the mole ratio of $\mathrm{Al}$ and $\mathrm{Cl}{\phantom{A}}_{2}$ required by balanced equation is
$\frac{\text{moles of Al}}{{\text{moles of Cl}}_{2}}\text{(required)}=\frac{2}{3}=0.6\stackrel{―}{6}$
and the actual mole ratio is
$\frac{\text{moles of Al}}{{\text{moles of Cl}}_{2}}\text{(actual)}=\frac{1.04×{10}^{-1}}{5.85×{10}^{-2}}=1.78$
Since the actual ratio is greater than the required ratio, we have more $\text{Al}$ than is needed to completely react the $\mathrm{Cl}{\phantom{A}}_{2}$. This means that the ${\text{Cl}}_{2}$ must be the limiting reactant. If the actual ratio had been smaller than the required ratio, then we would have had excess ${\text{Cl}}_{2}$, instead, and the $\text{Al}$ would be limiting.
Method 2: For the second method, we'll use the mole ratio between $\mathrm{Al}$ and $\mathrm{Cl}{\phantom{A}}_{2}$ to determine how much $\mathrm{Cl}{\phantom{A}}_{2}$ we would need to fully consume $1.04×{10}^{-1}$ moles of $\mathrm{Al}$. Then, we'll compare the answer to the amount of $\mathrm{Cl}{\phantom{A}}_{2}$ we actually have to see if $\mathrm{Cl}{\phantom{A}}_{2}$ is limiting or not. The number of moles of $\mathrm{Cl}{\phantom{A}}_{2}$ required to react with $1.04×{10}^{-1}$ moles of $\mathrm{Al}$ is
$1.04×{10}^{-1}\phantom{\rule{0.278em}{0ex}}\overline{)\text{mol Al}}×\frac{3\phantom{\rule{0.278em}{0ex}}{\text{mol Cl}}_{2}}{2\phantom{\rule{0.278em}{0ex}}\overline{)\text{mol Al}}}=1.56×{10}^{-1}\phantom{\rule{0.278em}{0ex}}{\text{mol Cl}}_{2}$
According to our earlier calculations, we have $5.85×{10}^{-2}$ moles of $\mathrm{Cl}{\phantom{A}}_{2}$, which is less than $1.56×{10}^{-1}$ moles. Again, this means that the $\mathrm{Cl}{\phantom{A}}_{2}$ is limiting. (Note that we could have done a similar analysis for $\mathrm{Al}$ instead of $\mathrm{Cl}{\phantom{A}}_{2}$, and we would have arrived at the same conclusion.)
Method 3: For the third and final method, we'll use mole ratios from the balanced equation to calculate the amount of $\mathrm{AlCl}{\phantom{A}}_{3}$ that would be formed by complete consumption of $\mathrm{Al}$ and $\mathrm{Cl}{\phantom{A}}_{2}$. The reactant that produces the smallest amount of $\mathrm{AlCl}{\phantom{A}}_{3}$ must be limiting. To start, let's calculate how much $\mathrm{AlCl}{\phantom{A}}_{3}$ would be formed if the $\mathrm{Al}$ was completely consumed:
$1.04×{10}^{-1}\phantom{\rule{0.278em}{0ex}}\overline{)\text{mol Al}}×\frac{2\phantom{\rule{0.278em}{0ex}}{\text{mol AlCl}}_{3}}{2\phantom{\rule{0.278em}{0ex}}\overline{)\text{mol Al}}}=1.04×{10}^{-1}\phantom{\rule{0.278em}{0ex}}{\text{mol AlCl}}_{3}$
Then, let's calculate the amount of $\mathrm{AlCl}{\phantom{A}}_{3}$ that would be formed if the $\mathrm{Cl}{\phantom{A}}_{2}$ was completely consumed:
$5.85×{10}^{-2}\phantom{\rule{0.278em}{0ex}}\overline{){\text{mol Cl}}_{2}}×\frac{2\phantom{\rule{0.278em}{0ex}}{\text{mol AlCl}}_{3}}{3\phantom{\rule{0.278em}{0ex}}\overline{){\text{mol Cl}}_{2}}}=3.90×{10}^{-2}\phantom{\rule{0.278em}{0ex}}{\text{mol AlCl}}_{3}$
Since the $\mathrm{Cl}{\phantom{A}}_{2}$ produces a smaller amount of $\mathrm{AlCl}{\phantom{A}}_{3}$ than the $\mathrm{Al}$ does, the $\mathrm{Cl}{\phantom{A}}_{2}$ must be the limiting reactant.

### Step 3: Calculate the theoretical yield

Our final step is to determine the theoretical yield of ${\mathrm{AlCl}}_{3}$ in the reaction. Remember that the theoretical yield is the amount of product that is produced when the limiting reactant is fully consumed. In this case, the limiting reactant is $\mathrm{Cl}{\phantom{A}}_{2}$, so the maximum amount of ${\mathrm{AlCl}}_{3}$ that can be formed is
$5.85×{10}^{-2}\phantom{\rule{0.278em}{0ex}}\overline{){\text{mol Cl}}_{2}}×\frac{2\phantom{\rule{0.278em}{0ex}}{\text{mol AlCl}}_{3}}{3\phantom{\rule{0.278em}{0ex}}\overline{){\text{mol Cl}}_{2}}}=3.90×{10}^{-2}\phantom{\rule{0.278em}{0ex}}{\text{mol AlCl}}_{3}$
Note that we had already calculated this value while working through Method 3! Since a theoretical yield is typically reported with units of mass, let's use the molar mass of $\mathrm{AlCl}{\phantom{A}}_{3}$ to convert from moles of $\mathrm{AlCl}{\phantom{A}}_{3}$ to grams:
$3.90×{10}^{-2}\phantom{\rule{0.278em}{0ex}}\overline{){\text{mol AlCl}}_{3}}×\frac{133.33\phantom{\rule{0.278em}{0ex}}{\text{g AlCl}}_{3}}{1\phantom{\rule{0.278em}{0ex}}\overline{){\text{mol AlCl}}_{3}}}=5.20\phantom{\rule{0.278em}{0ex}}{\text{g AlCl}}_{3}$

## Percent yield

As we just learned, the theoretical yield is the maximum amount of product that can be formed in a chemical reaction based on the amount of limiting reactant. In practice, however, the actual yield of product—the amount of product that is actually obtained—is almost always lower than the theoretical yield. This can be due to a number of factors, including side reactions (secondary reactions that form undesired products) or purification steps that lower the amount of product isolated after the reaction.
The actual yield of a reaction is typically reported as a percent yield, or the percentage of the theoretical yield that was actually obtained. The percent yield is calculated as follows:
$\text{Percent yield}=\frac{\text{actual yield}}{\text{theoretical yield}}×100\mathrm{%}$
Based on this definition, we would expect a percent yield to have a value between 0% and 100%. If our percent yield is greater than 100%, that means we probably calculated something incorrectly or made an experimental error. With all this in mind, let's try calculating the percent yield for a precipitation reaction in the following example.

## Example 2: Calculating percent yield

A students mixes of $0.314\phantom{\rule{0.278em}{0ex}}M$ $\mathrm{BaCl}{\phantom{A}}_{2}$ with excess $\mathrm{AgNO}{\phantom{A}}_{3}$, causing $\mathrm{AgCl}$ to precipitate. The balanced equation for the reaction is shown below.
$\mathrm{BaCl}{\phantom{A}}_{2}\left(aq\right)+2\phantom{\rule{0.167em}{0ex}}\mathrm{AgNO}{\phantom{A}}_{3}\left(aq\right)\to 2\phantom{\rule{0.167em}{0ex}}\mathrm{AgCl}\left(s\right)+\mathrm{Ba}\left(\mathrm{NO}{\phantom{A}}_{3}\right){\phantom{A}}_{2}\left(aq\right)$
If the student isolates of $\mathrm{AgCl}\left(s\right)$, what is the percent yield of the reaction?
To solve this problem, we'll need to use the given information about the limiting reactant, $\mathrm{BaCl}{\phantom{A}}_{2}$, to calculate the theoretical yield of $\mathrm{AgCl}$ for the reaction. Then, we can compare this value to the actual yield of $\mathrm{AgCl}$ to determine the percent yield. Let's walk through the steps now:

### Step 1: Find moles of the limiting reactant

To determine the theoretical yield of $\mathrm{AgCl}$, we first need to know how many moles of $\mathrm{BaCl}{\phantom{A}}_{2}$ were consumed in the reaction. We're given the volume () and molarity ($0.314\phantom{\rule{0.278em}{0ex}}M$) of the $\mathrm{BaCl}{\phantom{A}}_{2}$ solution, so we can find the number of moles of $\mathrm{BaCl}{\phantom{A}}_{2}$ by multiplying these two values:
$\begin{array}{rl}{\text{Moles BaCl}}_{2}& ={M}_{{\text{BaCl}}_{2}}×\text{liters soln}\\ \\ & =\frac{0.314\phantom{\rule{0.278em}{0ex}}{\text{mol BaCl}}_{2}}{1\phantom{\rule{0.278em}{0ex}}\overline{)\text{L soln}}}×.0250\phantom{\rule{0.278em}{0ex}}\overline{)\text{L soln}}\\ \\ & =7.85×{10}^{-3}\phantom{\rule{0.278em}{0ex}}{\text{mol BaCl}}_{2}\end{array}$

### Step 2: Determine the theoretical yield (in grams)

Now that we have the quantity of $\mathrm{BaCl}{\phantom{A}}_{2}$ in moles, we can find the theoretical yield of $\mathrm{AgCl}$ in grams:
$7.85×{10}^{-3}\phantom{\rule{0.278em}{0ex}}\overline{){\text{mol BaCl}}_{2}}×\frac{2\phantom{\rule{0.278em}{0ex}}\overline{)\text{mol AgCl}}}{1\phantom{\rule{0.278em}{0ex}}\overline{){\text{mol BaCl}}_{2}}}×\frac{143.32\phantom{\rule{0.278em}{0ex}}\text{g AgCl}}{1\phantom{\rule{0.278em}{0ex}}\overline{)\text{mol AgCl}}}=2.25\phantom{\rule{0.278em}{0ex}}\text{g AgCl}$

### Step 3: Calculate the percent yield

Finally, we can calculate the percent yield of $\mathrm{AgCl}$ by dividing the actual yield of $\mathrm{AgCl}$ () by our theoretical yield from Step 2:
$\begin{array}{rl}\mathrm{%}\phantom{\rule{0.278em}{0ex}}\text{yield}& =\frac{\text{actual yield}}{\text{theoretical yield}}×100\mathrm{%}\\ \\ & =\frac{1.82\phantom{\rule{0.278em}{0ex}}\text{g AgCl}}{2.25\phantom{\rule{0.278em}{0ex}}\text{g AgCl}}×100\mathrm{%}\\ \\ & =80.9\mathrm{%}\end{array}$

### Summary

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed. As we saw in Example 1, there are many different ways to determine the limiting reactant, but they all involve using mole ratios from the balanced chemical equation.
The amount of product that can be formed based on the limiting reactant is called the theoretical yield. In reality, the amount of product actually collected, known as the actual yield, is almost always smaller than the theoretical yield. The actual yield is usually expressed as a percent yield, which specifies what percentage of the theoretical yield was obtained.

## Want to join the conversation?

• It said that if you get a percent yield over 100 it'sometimes because you made a mistake in the lab, which makes sense to me, but I've heard that there can be other, more scientific reasons.... how does that work?
• Nice question! The theoretical yield is maximum 100% by definition. You can run an experiment correctly and still get >100% although It's not quite a "mistake" as you say: your final product will often include impurities (things other than your pure desired compound) so it will seem like you made more than you could have gotten because you also weigh the impurities. The other, simillar, reason is that your final product is not completely dry, in which the "impurity" would be water.
• In Step 3, Calculation of Percent Yield the equation shows 1.82/2.15 as equaling 83.9%. I believe it should say 84.7%, If I'm incorrect in this I blame Texas Instruments the producer of my calculator.
• You are correct. The answer is not 83.9, but 84.6.
It's 84.6 instead of 84.7 because of minor rounding issues.

Kudos to Texas Instruments.
• Where do you get the actual yield from? Thanks:)
• If speaking in terms of doing a lab or experiment, the actual yield comes from the actual result of the lab (hence the name).
• On a standardized test, how would you distinguish differences between a solely stoichiometric problem and a limiting reagent problem?
• The amount of the reactants would be given to you.
• How would you express the actual yield if a side reaction occurs?
• In general, the theoretical yield is calculated assuming no side reactions will occur (this is almost never actually the case! But still that is the assumption that is usually made). In the case that there is a side reaction, you would calculate the actual yield based on how much pure product you were able to make. This means you need to be able to either separate your product from the side products (or leftover reactants), or you have some analytical method to analyze the purity of your product.
• For the percent yield equation, must the equation be in grams or can it be done in moles as well?
• You can use whatever units you wish, provided the actual yield and the theoretical yield are expressed in the same units.
• Could anyone give more detail about why the actual yield is almost always lower than theoretical? What exactly is meant by side reactions and purification steps?
• The theoretical yield assumes that all of your reactants (100% of them) react together in the desired reaction to produce your products. But real life is more messy than idealized math solutions and mistakes happen. Your actual yield is almost always going to be less than your theoretical yield because you do not obtain the entirety of your product.

One reason is that you can simply have unreacted reactants which do not produce products. This can be remedied by increasing the contact reactants have with each other such as by better (or constant) stirring.

Another reason if that often you are transferring solutions to and from glassware for entire reactions and products can simply be spilled by accident. Or they can be left in the glassware attached to tiny scratches on the inside of the glassware where you are unable to collect it. This is mendable by having good quality glassware and being careful as a chemist when transferring solutions.

Another common frustration is the occurrence of side reactions, or undesired reactions which creates a product that you did not wish to have. This can actually create a scenario where your percent yield is actually higher than your theoretical yield because it is contaminated with impurities. These can possibly be fixed by being extra cautious to not expose your reaction to air since chemicals like water vapor in the air can react with your reaction to create impurities.

That being said if you performed your reaction well and produced a pure sample your percent yield will be less than 100%. An excellent percent yield would be somewhere in the 90% range. 80% is considered good. Around 50% is considered adequate.

Hope that helps.
• So there isn't a way to find the actual yield without doing an experiment?
• Theoretical yield is what you think should happen:,
and is a calculated amount derived through stoichiometry.
Actual yield is what you observed to happen or you have been told happened:
and is derived by measurement in an experiment or manufacturing process.
Expected yield is the amount that has been consistently reported, when control measures are used.
So there is another way to figure yield as expected yield, which would include the parameters and control methods used to increase yield in a real life manufacture of a chemical process.
We can use theoretical yield or expected yield to calculate percent yield.