Limiting reactant and reaction yields (article)

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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=1.04×10−1mol Al4.15g Cl2×1mol Cl270.90g Cl2=5.85×10−2mol 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)=1.04×10−15.85×10−2=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 1.04×10−1 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 1.04×10−1 moles of Al is

1.04×10−1mol Al×3mol Cl22mol Al=1.56×10−1mol Cl2

According to our earlier calculations, we have 5.85×10−2 moles of ClA2, which is less than 1.56×10−1 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:

1.04×10−1mol Al×2mol AlCl32mol Al=1.04×10−1mol AlCl3

Then, let’s calculate the amount of AlClA3 that would be formed if the ClA2 was completely consumed:

5.85×10−2mol Cl2×2mol AlCl33mol Cl2=3.90×10−2mol 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

5.85×10−2mol Cl2×2mol AlCl33mol Cl2=3.90×10−2mol 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:

3.90×10−2mol AlCl3×133.33g AlCl31mol AlCl3=5.20g AlCl3



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