Chapter 11: Stoichiometry
Sec. 11.3: Limiting Reactants
Objectives
 Identify the limiting reactant in a chemical
equation.
 Identify the excess reactant and calculate
the amount remaining after the reaction is
complete.
 Calculate the mass of a product when the
amounts of more than one reactant are
given.
Limiting Reactants
If I want to make
s’mores, does it
matter how many
marshmallows I have if
I only have one piece of
chocolate?
No!! I will use up the
chocolate & there will
be marshmallows left
over!
Limiting Reactants – p. 379, Fig. 4
 If I have 10 screwdrivers, 5 pliers, & 4 hammers,
how many tool sets could I make?
 Each tool set consists of 2 screwdrivers, a pliers
and a hammer.
 The hammers are used up. The number of
hammers limits how many sets can be made.
 There are leftover pliers and screwdrivers. There
is an excess of these tools.
In Chemical Reactions:
 A reaction will proceed until one reactant is
used up.
 The amount of product formed depends
upon the reactant that is limited. When this
reactant is used up, the reaction stops.
 The limiting reactant is the reactant that
limits the extent of the reaction and
determines the amount of product formed.
In Chemical Reactions:
 When the limiting reactant is used up, a
portion of all of the other reactants will
remain after the reaction stops.
 These left-over reactants are called excess
reactants.
Consider this reaction:
3H2 + 3N2 --> 2NH3 +
2N2
Visualize all the nitrogen and hydrogen atoms separating.
The atoms are then available to reassemble into molecules
(like the tools are assembled into kits.) Only two ammonia
molecules can be assembled because there are 6 hydrogen
atoms and each ammonia molecule requires 3!
When the hydrogen is gone, some nitrogen atoms remain
unreacted. Hydrogen is limiting and nitrogen is in excess.
Determining the limiting reactant &
amount of product
 If 200g of sulfur reacts with 100 g of
chlorine, what mass of disulfur
dichloride is produced?
S8(l) + 4 Cl2(g) → 4S2Cl2(l)
Remember: The amount of product
depends on the reactant that is limited.
Finding that is your first step!
S8(l) + 4 Cl2(g) → 4S2Cl2(l)
 From the masses of reactants given,
determine the number of moles of reactants


100 g Cl2 x 1 mol = 1.41 mol Cl2
71 g
200 g S8 x 1 mol = 0.779 mol S8
256.8 g
S8(l) + 4 Cl2(g) → 4S2Cl2(l)
 Next, determine whether the 2 reactants are
in the correct mole ratio. The equation
indicates that the ratio of chlorine to sulfur
is 4:1. Find the actual mole ratio by
dividing the mole values you just
calculated.

1.41 mol Cl2 = 1.81 mol Cl2
0.779 mol S8
mol S8
The Limiting Reactant



This means that 1.81 mol of chlorine is
available for each mole of S8.
The mole ratio says we need 4 chlorine
for 1 sulfur.
There is not enough chlorine to use up all
the sulfur so chlorine is the limiting
reactant!!
Practice Problems
 Identify the limiting reactant when 3.50 g of
HCl reacts with 5.28 g of NaOH to produce
NaCl and water.
 Identify the limiting reactant when 1.22 g
O2 reacts with 1.05g H2 to produce water.
Determining the amount of product
 Since the limiting reactant determines the
amount of product formed, the amount of
product can be determined when the
limiting reactant is known.
 The moles of the limiting reactant will be
our “known” in a standard stoichiometric
calculation. The mass of the product is the
unknown.
Let’s use the example from before. We determined that
Cl2 was limiting and that there were 1.41 mol of it.
S8(l) + 4 Cl2(g) → 4S2Cl2(l)
1.41 mol

1.41 mol Cl2 x 4 mol S2Cl2
4 mol Cl2
x 135.2 g = 191 g S2Cl2
1 mol
? Grams
Practice Problems
 Determine the mass of tetraphosphorus
decoxide formed if 25.0 g of phosphorus
(P4) and 50.0 g oxygen are combined.
P4 + 5O2 --> P4O10
 In photosynthesis, 6CO2 + 6H2O -->
C6H12O6 + 6O2. If a plant has 88.0 g of
CO2 and 64.0 g H2O available, what mass
of glucose will be produced?
Excess Reactants
 Once the limiting reactant has been
determined, it is also possible to use
stoichiometry to find out how much of the
excess reactant is used or how much is
leftover.
 Recall this reaction:
S8(l) + 4 Cl2(g) → 4S2Cl2(l)
We determined the limiting reactant is
chlorine, with the amount of 1.41 moles.
S8(l) + 4 Cl2(g) → 4S2Cl2(l)
• With the known of 1.41 mol chlorine, we can
determine the no. of moles and grams of sulfur
that will be used:
1.41 mol Cl2 x 1 mol S8 = 0.353 mol S8
4 mol Cl2
0.353 mol S8 x 256.8 g = 90.7 g S8 USED
1 mol
• Subtract to get the amount that is left over:
200 g (the original amount of S8) – 90.7 g =
109.3 g LEFTOVER
Practice Problems
1. What mass of SO3 is produced from the
reaction of 10.5 g of SO2 and 4.32 g of O2?
How much of the excess reactant is left
when the reaction is complete?
2. If 8.0 g of Cr is heated with 16.7 g of Cl2,
what mass of CrCl3 will be produced?
What is the excess reactant and how much
of it is used in the reaction?