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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 2.80 g sample of Al(s) reacts with a 4.15 g sample of ClA2(g) according to the equation shown below.
2Al(s)+3ClA2(g)2AlClA3(s)
What is the theoretical yield of AlClA3 in this reaction?
To solve this problem, we first need to determine which reactant, Al or ClA2, 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 AlClA3.

Step 1: Convert reactant masses to moles

Let's start by converting the masses of Al and ClA2 to moles using their molar masses:
2.80g Al×1mol Al26.98g Al=1.04×101mol Al4.15g Cl2×1mol Cl270.90g Cl2=5.85×102mol Cl2

Step 2: Find the limiting reactant

Now that we know the quantities of Al and ClA2 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 Al and ClA2 in the balanced equation to the mole ratio actually present. In this case, the mole ratio of Al and ClA2 required by balanced equation is
moles of Almoles of Cl2(required)=23=0.66
and the actual mole ratio is
moles of Almoles of Cl2(actual)=1.04×1015.85×102=1.78
Since the actual ratio is greater than the required ratio, we have more Al than is needed to completely react the ClA2. This means that the Cl2 must be the limiting reactant. If the actual ratio had been smaller than the required ratio, then we would have had excess Cl2, instead, and the Al would be limiting.
Method 2: For the second method, we'll use the mole ratio between Al and ClA2 to determine how much ClA2 we would need to fully consume 1.04×101 moles of Al. Then, we'll compare the answer to the amount of ClA2 we actually have to see if ClA2 is limiting or not. The number of moles of ClA2 required to react with 1.04×101 moles of Al is
1.04×101mol Al×3mol Cl22mol Al=1.56×101mol Cl2
According to our earlier calculations, we have 5.85×102 moles of ClA2, which is less than 1.56×101 moles. Again, this means that the ClA2 is limiting. (Note that we could have done a similar analysis for Al instead of ClA2, 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 AlClA3 that would be formed by complete consumption of Al and ClA2. The reactant that produces the smallest amount of AlClA3 must be limiting. To start, let's calculate how much AlClA3 would be formed if the Al was completely consumed:
1.04×101mol Al×2mol AlCl32mol Al=1.04×101mol AlCl3
Then, let's calculate the amount of AlClA3 that would be formed if the ClA2 was completely consumed:
5.85×102mol Cl2×2mol AlCl33mol Cl2=3.90×102mol AlCl3
Since the ClA2 produces a smaller amount of AlClA3 than the Al does, the ClA2 must be the limiting reactant.

Step 3: Calculate the theoretical yield

Our final step is to determine the theoretical yield of AlCl3 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 ClA2, so the maximum amount of AlCl3 that can be formed is
5.85×102mol Cl2×2mol AlCl33mol Cl2=3.90×102mol AlCl3
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 AlClA3 to convert from moles of AlClA3 to grams:
3.90×102mol AlCl3×133.33g AlCl31mol AlCl3=5.20g AlCl3

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.
Oh no, a cat burglar stole one of our hot dog buns! This means that our actual yield is only three complete hot dogs. Given that the theoretical yield was four complete hot dogs, what is our percent yield?
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:
Percent yield=actual yieldtheoretical yield×100%
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 25.0 mL of 0.314M BaClA2 with excess AgNOA3, causing AgCl to precipitate. The balanced equation for the reaction is shown below.
BaClA2(aq)+2AgNOA3(aq)2AgCl(s)+Ba(NOA3)A2(aq)
If the student isolates 1.82 g of AgCl(s), what is the percent yield of the reaction?
To solve this problem, we'll need to use the given information about the limiting reactant, BaClA2, to calculate the theoretical yield of AgCl for the reaction. Then, we can compare this value to the actual yield of 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 AgCl, we first need to know how many moles of BaClA2 were consumed in the reaction. We're given the volume (0.0250 L) and molarity (0.314M) of the BaClA2 solution, so we can find the number of moles of BaClA2 by multiplying these two values:
Moles BaCl2=MBaCl2×liters soln=0.314mol BaCl21L soln×.0250L soln=7.85×103mol BaCl2

Step 2: Determine the theoretical yield (in grams)

Now that we have the quantity of BaClA2 in moles, we can find the theoretical yield of AgCl in grams:
7.85×103mol BaCl2×2mol AgCl1mol BaCl2×143.32g AgCl1mol AgCl=2.25g AgCl

Step 3: Calculate the percent yield

Finally, we can calculate the percent yield of AgCl by dividing the actual yield of AgCl (1.82 g) by our theoretical yield from Step 2:
%yield=actual yieldtheoretical yield×100%=1.82g AgCl2.25g AgCl×100%=80.9%

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?

  • leafers tree style avatar for user micah.ariel.snow
    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?
    (53 votes)
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    • old spice man green style avatar for user Matt B
      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.
      (123 votes)
  • leafers ultimate style avatar for user Jonny Cartee
    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.
    (30 votes)
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  • leaf green style avatar for user brookeleath
    Where do you get the actual yield from? Thanks:)
    (13 votes)
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  • piceratops ultimate style avatar for user SpamShield2.0
    On a standardized test, how would you distinguish differences between a solely stoichiometric problem and a limiting reagent problem?
    (7 votes)
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  • piceratops ultimate style avatar for user SpamShield2.0
    How would you express the actual yield if a side reaction occurs?
    (4 votes)
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    • blobby green style avatar for user yuki
      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.
      (15 votes)
  • duskpin ultimate style avatar for user kathrynjade777
    For the percent yield equation, must the equation be in grams or can it be done in moles as well?
    (7 votes)
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  • blobby green style avatar for user waissene
    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?
    (4 votes)
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    • leaf red style avatar for user Richard
      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.
      (9 votes)
  • winston default style avatar for user Gray
    So there isn't a way to find the actual yield without doing an experiment?
    (6 votes)
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    • blobby green style avatar for user George Karkanis
      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.
      (3 votes)
  • blobby green style avatar for user Emma  Salgado
    In step 2 method 1, how did you go from having 1.74 mol of Al to 0.67 mol of Al?
    (4 votes)
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  • duskpin tree style avatar for user Deblina
    When calculating theoretical yield with an acid, does the molarity of the acid affect the calculations?
    (5 votes)
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