SCH3U
5.5 Calculating Limiting and Excess Reagents
The quantities of substances that react in a chemical reaction are determined by the
mole ratio, but in the real world, it is less likely for the reactants to be in the mole
ratios given by the equation.
Why?
 In nature, compounds are found in random quantities
 In labs/industry, we might want to ensure that one reactant is completely
used up (eg. neutralizing toxins)
Limiting reagent:
o the reagent in a chemical reaction that limits the amount of product that is
able to form
o it is completely used up in the reaction.
Excess reagent:
o the reagent that is not completely used up in the reaction
o there will be some left over at the end of the reaction
o in order to provide a reagent in excess, chemists add 10% to the mass
required
Kitchen Stoichiometry – An Analogy
A recipe to make 30 strawberry candies requires the following:
2 cups water
5 cups sugar
1 cup strawberries
Equation:
2 cups water + 5 cups sugar + 1 cup strawberries 30 strawberry candies
Questions:
a. If you were given 10 cups of water, 10 cups of sugar and 10 cups of
strawberries, which ingredient would limit the amount of candies being
made? Which reactant will be in excess?
10 cups of water would make 150 candies; 10 cups of sugar would make 60
candies; 10 cups of strawberries would make 300 candies. Therefore, sugar
would limit the amount of candies being made.
Therefore, water and strawberries will be in excess (left over).
b. How much candies can you make with the ingredients given in part a?
10 cups of sugar X 30 candies
= 60 candies
5 cups of sugar
60 strawberry candies.
SCH3U
5.5 Calculating Limiting and Excess Reagents
c. How much of the excess reagents will be left over?
4 cups of water are needed for 10 cups of sugar. Therefore 6 cups (10 – 4)
water would be left over.
2 cups of strawberries are needed for 10 cups of sugar. Therefore 8 cups (10
-2) of strawberries would be left over.
How to determine which reactant is the Limiting Reagent
1. Write a balanced chemical equation.
2. Pick a product (choose the product that you will need to eventually
determine the amount of).
3. Determine the amount of the chosen product (in moles) that can form from
each of the reactants.
4. Compare the amount of products formed for all reactants – the reactant that
produces the least amount of the product is the limiting reagent.
Example 1:
Lithium nitride reacts with water to produce ammonia gas and lithium hydroxide. If
4.5g of lithium nitride and 5.8 g of water is available:
a. Which reactant is the limiting reagent and which is the excess reagent?
Li3N (S)
4.5g Li3N
5.8 g H2O
+
3H2O (l)
X
X

NH3 (g)
1 mol Li3N
34.83 g Li3N
1 mol H2O
18.02 g H2O
X
X
+
3LiOH (aq)
1 mol NH3 =
1 mol Li3N
1 mol NH3 =
3 mol H2O
0.13 mol NH3
0.11 mol NH3
5.8 g of water produces less NH3 than 4.5g of lithium nitrate. Therefore, water is the
limiting reagent and lithium nitride is the excess reagent.
b. How much (in grams) of ammonia gas can be produced?
(look at a – you must use the number of moles produced by the limiting
reagent)
0.11 mol NH3
X
17.04 g NH3
1 mol NH3
=
1.9 g NH3
c. How much (in grams) of the excess reagent will be left over?
0.11mol NH3
X
1 mol Li3N
X
34.83 g Li3N =
3.8 g Li3N
1 mol NH3
1 mol Li3N
3.8 g of lithium nitride is used up. Therefore, 0.7 g (4.5 – 3.8) is left over.
SCH3U
Example 2:
5.5 Calculating Limiting and Excess Reagents
Nitrogen and hydrogen react to form ammonia according to the following equation:
N2 (g)
+
3H2 (g)

2NH3 (g)
If 28.02g of N2 is to be completely used up, what is a reasonable mass of H2 to use in
excess in the reaction?