SCH 4C
Unit 2: Quantities in Chemical Reactions
Limiting and Excess Reagents and Percentage Yield
Term
Definition
Excess reagent
The reactant that is present in more than the required amount for a complete reaction to
occur.
Limiting reagent
The reactant that is completely used up in a chemical reaction. It is the information about
the limiting reagent that should be used to calculate how much product will be formed in
the reaction.
Theoretical yield
This is the quantity of a product that you predict should be formed based on how much
reactant you have and on the balanced chemical equation. If there is a limiting reagent,
you MUST use its information to calculate the theoretical yield of product.
Actual yield
The quantity of product that you actually produce in the lab during a chemical reaction.
This value is often different from the theoretical yield for a variety of reasons. A lower
actual yield than you predict with your theoretical yield may occur because of impurities in
the reactants, because of other reactions going on (side reactions) that form different
products from those you are interested in, or because some reactions are reversible, so
some products may react together to form reactants.
Percentage yield
actual yield
percentage yield=
theoretical yield
x100%
Example
Table salt, NaCl(s), can be formed by the reaction of sodium metal and chlorine gas as shown below. A reaction
mixture is set up with 45.98 g of sodium and 142.0 g of chlorine gas.
2 Na(s) + Cl2(g)  2NaCl(s)
a) Identify which reactant is limiting and which one is in excess.
b) Calculate the mass of sodium chloride you would expect to produce when the reaction is over. This is the
theoretical yield of NaCl.
c) In a lab, 50.0 g of NaCl(s) were actually produced in this reaction. What is the percentage yield?
Solution:
Part a) Finding the limting reagent
2 Na(s) +
Mass Given
45.98 g
Molar Mass of Substances
g
22.99
Mentioned

142.0 g
Cl2(g)
Mole Ratio of Substances
Mentioned from the
balanced equation
Which reactant is limiting
and which is in excess?
m
mm
45.98 g

g
22.99
mol
 2.0 mol
?g
mmCl2  2(35.45)
mol
No. of Moles of the
substance(s) you can
calculate it for
2 NaCl(s)
 70.0
g
mol
mmNaCl  22.99  35.45
 58.44
m
mm
142.0 g

g
70.0
mol
 0.6 mol
nNa 
nCl2 
2 mol Na(s)
1 mol Cl2(g)
This reactant is in
excess because
we will have some
left over if only
0.6 mol of Cl2 is
available to react.
This reactant is
limiting. If we have
2 mol of Na, we
need 1 mol of Cl2
but we only have
0.6 mol of Cl2
available.
2 mol NaCl(s)
g
mol
Part b) Finding the theoretical yield
You must use the information about the limiting reagent to calculate the theoretical yield of a product. So – we are
only interested in using the information about Cl2 to determine how much NaCl will be formed.
2 Na(s) +
Mass Given
Molar Mass of Substances
Mentioned
Cl2(g)

142.0 g
mmCl2  2(35.45)
 70.0
No. of Moles of the
substance(s) you can
calculate it for
2 NaCl(s)
?g
mmNaCl  22.99  35.45
g
mol
 58.44
m
mm
142.0 g

g
70.0
mol
 0.6 mol
nCl2 
Mole Ratio of Substances
Mentioned from the
balanced equation
1 mol Cl2(g)
g
mol
?
2 mol NaCl(s)
To determine the number of moles of the substance you need, use this equation:
nunknown  nknown 
nunknown frombalanced equation
nknown frombalanced equation
Now that you have the number of moles of Al2O3(s) produced, use m  n(mm) t o determine the mass of NaCl.
Part c) Finding the percentage yield.
percentage yield=
actual yield
x100%
theoretical yield