# Finding the Limiting Reactant

In an ideal world, there would always be an equal number of hot dogs and hot dog buns, the peanut butter would be used up in the same sandwich that finishes off the jelly, and there would always be enough clean underwear to last until the weekend. In reality, these things happen only some of the time (if ever), and the rest of the time one thing runs out first, be it the buns, the jelly, or the briefs. The chemistry lab is just as much a part of the real world as everything else. When we mix chemicals together, very often one of them will run out before the other, and when that happens, the chemical reaction stops.

If you have studied stoichiometry, you already know how to calculate exactly how much of one chemical is needed to react with another chemical, using a balanced chemical equation. (If you aren’t sure how to do this, go read “An Introduction to Stoichiometry” here at Helium to get up to speed.) That stoichiometric amount would give you exactly the right amount of chemical for the reaction to continue until both reactants were used up. If the reactants are mixed in any other proportion, one will have some left over after the other is completely consumed. In chemistry, we call the chemical that is completely used up the “Limiting Reactant”, and the one which has extra the “Excess Reactant”. The limiting reactant limits how much of the product can be made, and so it is important to know which reactant it is. Then, we can use our knowledge of the limiting reactant to calculate (using stoichiometry) how much of each product will be made, and also how much of the excess reactant will actually be used.

Take a look at the following chemical equation. (Imagine that -> is an arrow, and the numbers immediately preceded by letters are subscripts, please.)

2 Al + 3 Br2 -> 2 AlBr3

Remember that the chemical equation functions as a recipe, saying that we need three bromines (Br2) per two aluminums (Al) we wish to react. (This is simply a ratio.) For some amounts of reactants, it will be obvious which reactant would be used up. Consider the following cases:

1 mole of aluminum and 95 moles of bromine

10 moles of aluminum and 10 moles of bromine

In the first case, the bromine is overwhelmingly in abundance, so aluminum is without question the limiting reactant.

In the second, the numbers are the same, but the ratio says we use more bromine (3) than aluminum (2), so the bromine gets used up first, and it is the limiting reactant.

Depending on your confidence in your mental math skills, you may trust your judgment when naming the limiting reactant. When you do want to make sure, there is an easy way to check. Simply divide the moles of each reactant by the coefficient for that chemical. Whichever result is smaller will mark the limiting reactant. (And the larger one is the excess reactant.) Try this one for an example:

7.19 moles of aluminum and 10.78 moles of bromine

aluminum: 7.19 / 2 = 3.595
bromine: 10.78 / 3 = 3.593

Yes, those numbers are very close, and did I choose them deliberately to make guessing the correct answer difficult. You can see now that one is slightly smaller than the other, and will be consumed first. The bromine is the limiting reactant for this example. The aluminum is the excess reactant, and will have a very small amount left over after the reaction.

Once you know the limiting reactant, you can use it to calculate how much of the excess reactant will actually be used, using regular stoichiometry. Knowing this, it is a simple matter of subtraction to see how much of the excess reactant will be left over after the reaction too.

Using the same chemical reaction, let us try to find out which chemical is the excess reactant, and how much of it will be left over. Start with 5.20 moles of aluminum and 8.10 moles of bromine.

2 Al + 3 Br2 -> 2 AlBr3

Start by identifying the limiting reactant.

aluminum: 5.20 / 2 = 2.60
bromine: 8.10 / 3 = 2.70

The aluminum is the limiting reactant, and the bromine is the excess reactant.

Use stoichiometry to see how much bromine is actually needed.

5.20 moles (of Al) x 3 / 2 = 7.8 moles (of Br)

7.8 moles of the bromine are used in the reaction.

Subtract the amount that is used from the amount present to find what will be left over.

8.10 moles 7.80 moles = 0.30 moles (of Br left over)

Using this method, we now know that:
1. All of the aluminum is used up (limiting reactant)
2. All but 0.30 moles of the bromine is used up (excess reactant)

This process will work for any limiting reactant or excess reactant problem you come across. Just remember to always work in moles, and to use the limiting reactant for your stoichiometric calculations. (If your amounts are not given in moles, there are conversion methods for grams, liters, and molarity in the Helium article “An Introduction to Stoichiometry”.

You may also wish to further your knowledge of stoichiometry with the Helium articles:
“An Introduction to Stoichiometry” and
“Understanding Percent Yield”.