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Limiting reactant and reaction yields

AP.Chem:
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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, colon, 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.
A reaction with five hot dogs and four hot dog buns reacting to give four complete hot dogs and one leftover hot dog. The hot dog buns are the limiting reagent, and the leftover single hot dog is the excess reagent. The four complete hot dogs are the theoretical yield.
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, point, 80, space, g sample of A, l, left parenthesis, s, right parenthesis reacts with a 4, point, 15, space, g sample of ClX2(g)\ce{Cl2}(g) according to the equation shown below.
2Al(s)+3ClX2(g)2AlClX3(s)\ce{2Al}(s) + \ce{3Cl2}(g) \rightarrow \ce{2AlCl3}(s)
What is the theoretical yield of AlClX3\ce{AlCl3} in this reaction?
To solve this problem, we first need to determine which reactant, A, l or ClX2\ce{Cl2}, 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 AlClX3\ce{AlCl3}.

Step 1: Convert reactant masses to moles

Let's start by converting the masses of A, l and ClX2\ce{Cl2} to moles using their molar masses:
2.80  g Al×1  mol Al26.98  g Al=1.04×101  mol Al4.15  g Cl2×1  mol Cl270.90  g Cl2=5.85×102  mol Cl2\begin{aligned}2.80\; \cancel{\text{g Al}} &\times \dfrac{1\; \text{mol Al}}{26.98\; \cancel{\text{g Al}}} = 1.04 \times 10^{-1}\; \text{mol Al} \\\\ 4.15\; \cancel{\text{g Cl}_2} &\times \dfrac{1\; \text{mol Cl}_2}{70.90\; \cancel{\text{g Cl}_2}} = 5.85 \times 10^{-2}\; \text {mol Cl}_2\end{aligned}

Step 2: Find the limiting reactant

Now that we know the quantities of A, l and ClX2\ce{Cl2} 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 A, l and ClX2\ce{Cl2} in the balanced equation to the mole ratio actually present. In this case, the mole ratio of A, l and ClX2\ce{Cl2} required by balanced equation is
start fraction, start text, m, o, l, e, s, space, o, f, space, A, l, end text, divided by, start text, m, o, l, e, s, space, o, f, space, C, l, end text, start subscript, 2, end subscript, end fraction, start text, left parenthesis, r, e, q, u, i, r, e, d, right parenthesis, end text, equals, start fraction, 2, divided by, 3, end fraction, equals, 0, point, 6, start overline, 6, end overline
and the actual mole ratio is
start fraction, start text, m, o, l, e, s, space, o, f, space, A, l, end text, divided by, start text, m, o, l, e, s, space, o, f, space, C, l, end text, start subscript, 2, end subscript, end fraction, start text, left parenthesis, a, c, t, u, a, l, right parenthesis, end text, equals, start fraction, 1, point, 04, times, 10, start superscript, minus, 1, end superscript, divided by, 5, point, 85, times, 10, start superscript, minus, 2, end superscript, end fraction, equals, 1, point, 78
Since the actual ratio is greater than the required ratio, we have more start text, A, l, end text than is needed to completely react the ClX2\ce{Cl2}. This means that the start text, C, l, end text, start subscript, 2, end subscript must be the limiting reactant. If the actual ratio had been smaller than the required ratio, then we would have had excess start text, C, l, end text, start subscript, 2, end subscript, instead, and the start text, A, l, end text would be limiting.
Method 2: For the second method, we'll use the mole ratio between A, l and ClX2\ce{Cl2} to determine how much ClX2\ce{Cl2} we would need to fully consume 1, point, 04, times, 10, start superscript, minus, 1, end superscript moles of A, l. Then, we'll compare the answer to the amount of ClX2\ce{Cl2} we actually have to see if ClX2\ce{Cl2} is limiting or not. The number of moles of ClX2\ce{Cl2} required to react with 1, point, 04, times, 10, start superscript, minus, 1, end superscript moles of A, l is
1, point, 04, times, 10, start superscript, minus, 1, end superscript, start cancel, start text, m, o, l, space, A, l, end text, end cancel, times, start fraction, 3, start text, m, o, l, space, C, l, end text, start subscript, 2, end subscript, divided by, 2, start cancel, start text, m, o, l, space, A, l, end text, end cancel, end fraction, equals, 1, point, 56, times, 10, start superscript, minus, 1, end superscript, start text, m, o, l, space, C, l, end text, start subscript, 2, end subscript
According to our earlier calculations, we have 5, point, 85, times, 10, start superscript, minus, 2, end superscript moles of ClX2\ce{Cl2}, which is less than 1, point, 56, times, 10, start superscript, minus, 1, end superscript moles. Again, this means that the ClX2\ce{Cl2} is limiting. (Note that we could have done a similar analysis for A, l instead of ClX2\ce{Cl2}, 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 AlClX3\ce{AlCl3} that would be formed by complete consumption of A, l and ClX2\ce{Cl2}. The reactant that produces the smallest amount of AlClX3\ce{AlCl3} must be limiting. To start, let's calculate how much AlClX3\ce{AlCl3} would be formed if the A, l was completely consumed:
1, point, 04, times, 10, start superscript, minus, 1, end superscript, start cancel, start text, m, o, l, space, A, l, end text, end cancel, times, start fraction, 2, start text, m, o, l, space, A, l, C, l, end text, start subscript, 3, end subscript, divided by, 2, start cancel, start text, m, o, l, space, A, l, end text, end cancel, end fraction, equals, 1, point, 04, times, 10, start superscript, minus, 1, end superscript, start text, m, o, l, space, A, l, C, l, end text, start subscript, 3, end subscript
Then, let's calculate the amount of AlClX3\ce{AlCl3} that would be formed if the ClX2\ce{Cl2} was completely consumed:
5, point, 85, times, 10, start superscript, minus, 2, end superscript, start cancel, start text, m, o, l, space, C, l, end text, start subscript, 2, end subscript, end cancel, times, start fraction, 2, start text, m, o, l, space, A, l, C, l, end text, start subscript, 3, end subscript, divided by, 3, start cancel, start text, m, o, l, space, C, l, end text, start subscript, 2, end subscript, end cancel, end fraction, equals, 3, point, 90, times, 10, start superscript, minus, 2, end superscript, start text, m, o, l, space, A, l, C, l, end text, start subscript, 3, end subscript
Since the ClX2\ce{Cl2} produces a smaller amount of AlClX3\ce{AlCl3} than the A, l does, the ClX2\ce{Cl2} must be the limiting reactant.

Step 3: Calculate the theoretical yield

Our final step is to determine the theoretical yield of A, l, C, l, start subscript, 3, end subscript 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 ClX2\ce{Cl2}, so the maximum amount of A, l, C, l, start subscript, 3, end subscript that can be formed is
5, point, 85, times, 10, start superscript, minus, 2, end superscript, start cancel, start text, m, o, l, space, C, l, end text, start subscript, 2, end subscript, end cancel, times, start fraction, 2, start text, m, o, l, space, A, l, C, l, end text, start subscript, 3, end subscript, divided by, 3, start cancel, start text, m, o, l, space, C, l, end text, start subscript, 2, end subscript, end cancel, end fraction, equals, 3, point, 90, times, 10, start superscript, minus, 2, end superscript, start text, m, o, l, space, A, l, C, l, end text, start subscript, 3, end subscript
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 AlClX3\ce{AlCl3} to convert from moles of AlClX3\ce{AlCl3} to grams:
3, point, 90, times, 10, start superscript, minus, 2, end superscript, start cancel, start text, m, o, l, space, A, l, C, l, end text, start subscript, 3, end subscript, end cancel, times, start fraction, 133, point, 33, start text, g, space, A, l, C, l, end text, start subscript, 3, end subscript, divided by, 1, start cancel, start text, m, o, l, space, A, l, C, l, end text, start subscript, 3, end subscript, end cancel, end fraction, equals, 5, point, 20, start text, g, space, A, l, C, l, end text, start subscript, 3, end subscript

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.
Burglar cat with stripes and eye mask holding a stolen hot dog bun.
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:
start text, P, e, r, c, e, n, t, space, y, i, e, l, d, end text, equals, start fraction, start text, a, c, t, u, a, l, space, y, i, e, l, d, end text, divided by, start text, t, h, e, o, r, e, t, i, c, a, l, space, y, i, e, l, d, end text, end fraction, times, 100, percent
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, point, 0, space, m, L of 0, point, 314, M BaClX2\ce{BaCl2} with excess AgNOX3\ce{AgNO3}, causing A, g, C, l to precipitate. The balanced equation for the reaction is shown below.
BaClX2(aq)+2AgNOX3(aq)2AgCl(s)+Ba(NOX3)X2(aq)\ce{BaCl2}(aq) + \ce{2AgNO3}(aq) \rightarrow \ce{2AgCl}(s) + \ce{Ba(NO3)2}(aq)
If the student isolates 1, point, 82, space, g of A, g, C, l, left parenthesis, s, right parenthesis, what is the percent yield of the reaction?
To solve this problem, we'll need to use the given information about the limiting reactant, BaClX2\ce{BaCl2}, to calculate the theoretical yield of A, g, C, l for the reaction. Then, we can compare this value to the actual yield of A, g, C, l 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 A, g, C, l, we first need to know how many moles of BaClX2\ce{BaCl2} were consumed in the reaction. We're given the volume (0, point, 0250, space, L) and molarity (0, point, 314, M) of the BaClX2\ce{BaCl2} solution, so we can find the number of moles of BaClX2\ce{BaCl2} by multiplying these two values:
Moles BaCl2=MBaCl2×liters soln=0.314  mol BaCl21  L soln×.0250  L soln=7.85×103  mol BaCl2\begin{aligned}\text{Moles BaCl}_2 &= M_{\text{BaCl}_2} \times \text{liters soln} \\\\ &= \dfrac{0.314\; \text{mol BaCl}_2}{1\; \cancel{\text{L soln}}} \times .0250\; \cancel{\text{L soln}} \\\\ &= 7.85 \times 10^{-3}\; \text{mol BaCl}_2\end{aligned}

Step 2: Determine the theoretical yield (in grams)

Now that we have the quantity of BaClX2\ce{BaCl2} in moles, we can find the theoretical yield of A, g, C, l in grams:
7, point, 85, times, 10, start superscript, minus, 3, end superscript, start cancel, start text, m, o, l, space, B, a, C, l, end text, start subscript, 2, end subscript, end cancel, times, start fraction, 2, start cancel, start text, m, o, l, space, A, g, C, l, end text, end cancel, divided by, 1, start cancel, start text, m, o, l, space, B, a, C, l, end text, start subscript, 2, end subscript, end cancel, end fraction, times, start fraction, 143, point, 32, start text, g, space, A, g, C, l, end text, divided by, 1, start cancel, start text, m, o, l, space, A, g, C, l, end text, end cancel, end fraction, equals, 2, point, 25, start text, g, space, A, g, C, l, end text

Step 3: Calculate the percent yield

Finally, we can calculate the percent yield of A, g, C, l by dividing the actual yield of A, g, C, l (1, point, 82, space, g) by our theoretical yield from Step 2:
%  yield=actual yieldtheoretical yield×100%=1.82  g AgCl2.25  g AgCl×100%=80.9%\begin{aligned} \%\;\text{yield} &= \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\\\ &= \dfrac{1.82\; \text{g AgCl}}{2.25\; \text{g AgCl}} \times 100\% \\\\ &= 80.9\% \end{aligned}

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?
    (49 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.
      (103 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.
    (27 votes)
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  • leaf green style avatar for user brookeleath
    Where do you get the actual yield from? Thanks:)
    (12 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?
    (5 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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  • 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?
    (6 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?
    (3 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.
      (6 votes)