Stoichiometry: Chemical Formulas and Equations

Limiting reactants


Limiting Reactants

Suppose you are a chef preparing a breakfast for a group of people, and are planning to cook French toast. You make French toast the way you have always made it: one egg for every three slices of toast. You never waiver from this recipe, because the French toast will turn out to be either too soggy or too dry (arguably, you are anal retentive). There are 8 eggs and 30 slices of bread in the pantry. Thus, you conclude that you will be able to make 24 slices of French toast and not one slice more.

This is a similar situation with chemical reactions in which one of the reactants is used up before the others - the reaction stops as soon as one of the reactants is consumed. For example, in the production of water from hydrogen and oxygen gas suppose we have 10 moles of H2 and 7 moles of O2.

Because the stoichiometry of the reaction is such that 1 mol of O2 2 moles of H2, the number of moles of O2 needed to react with all of the H2 is:

Thus, after all the hydrogen reactant has been consumed, there will be 2 moles of O2 reactant left.

The reactant that is completely consumed in a chemical reaction is called the limiting reactant (or limiting reagent) because it determines (or limits) the amount of product formed. In the example above, the H2 is the limiting reactant, and because the stoichiometry is 2H2 2H2O (i.e. H2 H2O), it limits the amount of product formed (H2O) to 10 moles. We actually have enough oxygen (O2) to form 14 moles of H2O (1O2 2H2O).

One approach to solving the question of which reactant is the limiting reactant (given an initial amount for each reactant) is to calculate the amount of product that could be formed from each amount of reactant, assuming all other reactants are available in unlimited quantities. In this case, the limiting reactant will be the one that produces the least amount of potential product.


Consider the following reaction:

Suppose that a solution containing 3.50 grams of Na3PO4 is mixed with a solution containing 6.40 grams of Ba(NO3)2. How many grams of Ba3(PO4)2 can be formed?

1. First we need to convert the grams of reactants into moles:

2. Now we need to define the stoichiometric ratios between the reactants and the product of interest (Ba3(PO4)2):

2 Na3PO4 Ba3(PO4)2

3 Ba (NO3)2 Ba3(PO4)2

3. We can now determine the moles of product that would be formed if reactant were to be consumed in its entirety during the course of the reaction:

4. The limiting reactant is the Ba (NO3)2 and we could thus make at most 0.0082 moles of the Ba3(PO4)2 product.

5. 0.0082 moles of the Ba3(PO4)2 product would be equal to:



Theoretical yields

The quantity of product that is calculated to form when all of the limiting reactant is consumed in a reaction is called the theoretical yield.

The amount of product actually obtained is called the actual yield.

Actual yield < Theoretical yield

for the following reasons:

The percent yield of a reaction relates the actual yield to the theoretical yield:

Percent yield = x 100

For example, in the previous exercise we calculated that 4.94 grams of Ba3(PO4)2 product should be formed. This is the theoretical yield. If the actual yield were 4.02 grams the percent yield would be:


1996 Michael Blaber