LIMITING REAGENT
Taking Stoichiometric conversions one step further
Limiting Reagent
¨ 
The reactant that limits the amount of product that
can be formed.
¤  The
reaction will stop when all of the limiting reactant is
consumed
¨ 
The only way to know which reactant is limiting is to
compare the number of moles of each reactant
¤  Whichever
is smaller is the limiting reactant
Steps to Calculating Limiting Reagent
1. 
Convert each reactant listed into moles of the same
product
¤  The
2. 
smallest is the limiting reactant!
Use the limiting reactant to calculate the yield of
the product indicated in the questions
Example #1
¨ 
¨ 
Using the equation:
2Na + Cl2 à 2NaCl
Suppose that 6.70g of Na reacts with 3.20g of Cl2.
Which is the limiting reactant? How many grams of
NaCl will be produced?
Example #1
¨ 
Step 1: Calculate moles of reactant for each reactant
6.70 g Na
X
1 mol Na
=
0.293 mol Na
=
0.045 mol Cl2
22.99 g Na
3.20 g Cl2
X
1 mol Cl2
70.9 g Cl2
¨ 
0.045 mol of Cl2 is smaller than 0.293 mol of Na
¤ 
Therefore, Cl2 is the limiting reagent
Example #1
¨ 
¨ 
Using the equation:
2Na + Cl2 à 2NaCl
Suppose that 6.70g of Na reacts with 3.20g of Cl2.
Which is the limiting reactant? How many grams of
NaCl will be produced?
.045 mol
Cl2
X
2 mol NaCl
58.44 g NaCl
=
X
1 mol Cl2
1 mol NaCl
5.26 g NaCl
Example #2
¨ 
¨ 
Suppose that 80.0g of copper reacted with
25.0g of sulfur. What is the limiting reactant?
What is the maximum amount of copper (I)
sulfide that can be produced?
This problem is slightly different in that you need to
write the balanced equation!
2Cu + S à Cu2S
Example #2
¨ 
Step 1: Calculate moles of product from each
reactant
80.0 g Cu
X
1 mol Cu
=
1.26 mol Cu
=
0.78 mol S
63.55 g Cu
25.0 g S
X
1 mol Cu
32.07 g S
¨ 
Limiting Reagent: Sulfur
Example #2
¨ 
Suppose that 80.0g of copper reacted with
25.0g of sulfur. What is the limiting reactant?
What is the maximum amount of copper (I) sulfide
that can be produced?
.78 mol S
X
1 mol Cu2S
1 mol S
159.17 g Cu2S
=
X
1 mol Cu2S
124.15 g Cu2S
Leftovers
¨ 
¨ 
Sometimes you will be asked to determine how
much of the excess reactant you’ll have left
For excess, you will need an additional 2 steps
1. 
Calculate the amount of the actual reactant needed
using the theoretical yield calculated from the limiting
reactant
2. 
Subtract the difference of what was needed in the
excess reactant and the amount actually in the problem
Example #3
¨ 
¨ 
Suppose 25.5g of Calcium carbide, CaC2,
reacts with 15.5g of water. What is the
maximum amount of acetylene, C2H2, that can
be produced?
How much of the excess reactant will you
have left over?
CaC2 + 2H2O à C2H2 + Ca(OH)2
Example #3
¨ 
Step 1: Determine the limiting reagent
25.5 g CaC2
X
1 mol CaC2
=
0.398 mol CaC2
=
0.86 mol H2O
64.10 g CaC2
15.5 g H2O
X
1 mol H2O
18.01 g H2O
¨ 
Limiting reagent is CaC2 and Excess reagent is H2O
Example #3
¨ 
¨ 
Suppose 25.5g of Calcium carbide, CaC2, reacts
with 15.5g of water. What is the maximum
amount of acetylene, C2H2, that can be
produced?
How much of the excess reactant will you have
left over?
Example #3
¨ 
Step 2: Convert moles of limiting reagent to grams
of leftover reagent
.398 mol
CaC2
X
2 mol H2O
1 mol CaC2
X
18.01 g H2O
1 mol H2O
=
14.34 g H2O
Example #3
¨ 
This means that we will only make 14.34 g of H2O.
We had 15.5 g. So how much do we have left?
15.5 g -
Starting
Amount
14.34
g
=
Amount
Needed
1.16 g H2O Left
Percent Yield
¨ 
So far, every problem we have calculated with
stoichiometry has used a theoretical value
¤  This
¨ 
¨ 
is what we SHOULD get under ideal conditions
However, in the lab we don’t usually get that
amount L
In order to know how successful the experiment was,
we need to account for the amount that we got
versus what we should have gotten
Percent Yield
¨ 
Percent yield:
¤  The
ratio of the actual yield to the theoretical yield
¤  Expressed as a percentage
¨ 
% Yield = Actual Yield X 100
Theoretical Yield
Percent Yield Problems
¨ 
Must tell you two out of 3 pieces of information:
¤  The
amount of at least one reactant
¤  The amount of product that was produced
¤  The percent yield
¨ 
You will need to use the amount of the reactant
given to calculate the theoretical yield, then plug
into the % yield equation
Percent Yield Example #1
When 29.5g of CaCO3 is heated it decomposes into
CaO and CO2. 14.2g of CaO is actually produced.
What is the percent yield of CaO in this reaction?
¨  CaCO3 à CaO + CO2
¨ 
29.5g CaCO3 x 1 mol CaCO3 x 1 mol CaO x 56.08 g CaO =
100.09g CaCO3 1 mol CaCO3 1 mol CaO
14.2g CaO x 100% = 85.90%
16.53g CaO
16.53g
CaO