Section 8.3 pg. 320-324
 In any chemical reaction, it is easy to run out of one or
another reactant – which has an impact on the amount of
products that can result from a reaction
 To solve these problems you must identify which of the
reactants is going to run out first.
 This is the “limiting reagent”
 The other is the “excess reagent”
 Example: LEGO – If you had
three yellow blocks and four red blocks,
you could only make three yellow/red combinations – because there
is not enough yellow blocks to make four. The yellow block is the
limiting reagent and the red block is the excess reagent
+
=
 WHY DO WE CARE??
 It is often desirable to know how much excess reagent is
required to ensure that a reaction goes to completion.
 The general rule is to assume that a reasonable quantity of
excess reagent is to use 10% more than the quantity
required (Not the case in commercial chemistry)
 When you know the quantity of more than one reagent,
you may need to know which one will limit the reaction.
1)
You want to test the stoichiometric method using the
reaction of 2.00 g of copper(II) sulfate in solution with an
excess of sodium hydroxide in solution. What would be a
reasonable mass of sodium hydroxide to use?
 To answer this question, you need to calculate the minimum
mass required and then add 10%.
CuSO4(aq) + 2 NaOH(aq)  Cu(OH)2(s) + Na2SO4(aq)
2.00g
m=?
159.62g/mol 40.00g/mol
2.00 g x
1 mol x 2
159.62 g
1
1.00 g x 1.10%= 1.10g
x
40.0g
1 mol
= 1.00 g
10% = 0.10g
Practice pg. 321 #2
3)
If 10.0 mol of methane and 10.0 mol of oxygen react,
which is the limiting reagent.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
10.0 mol
10.0 mol
n CH4(g): 10.0 mol x 2
1
n O2(g): 10.0 mol x 1
2
= 20.o mol
= 5.00 mol
20.0 mol of oxygen
would be required
to react with 10.0
mol of methane
5.00 mol of methane
would be required to
react with 10.0 mol
of oxygen
Since 20.0 mol of oxygen is required to react with 10.0 mol of methane,
but only 10.0 mol is available, oxygen is the limiting reagent.
How much methane would be left? 5.0 mol – 5.00 mol = 5.00 mol
Practice pg. 324 #3b-d
2) If 10.0g of copper is placed in solution of 20.0g of silver
nitrate, which reagent will be the limiting reagent?
 All reactants must be converted to moles, then using the
mole ratio, determine which reactant will run out first.
Cu(s)
10.0g
63.55 g/mol
+ 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq)
20.0g
169.88g/mol
That much silver
nitrate is not
available so
copper is not the
limiting reagent
n Cu(s): 10.0g x 1 mol
= 0.157 mol x 2
63.55 g
= 0.315 mol
1
n AgNO3: 20.0g x 1 mol
= 0.118 mol x 1
169.88 g
= 0.0589 mol
2
You must test one of the values using the mole ratio.
Assume one chemical is completely used up and see
if enough of the second chemical is present.
More copper than
that is available so
silver nitrate is the
limiting reagent
3) From the previous example, where 10.0 g of copper reacts
with 20.0 g of silver nitrate, what mass of copper will be in
excess? (leftover when the reaction is complete)
ER
Cu(s)
0.157 mol
LR
+ 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq)
0.118 mol
n Cu(s): 0.118 mol x 1 = 0.0589 mol x 63.55 g = 3.74 g of copper
will be required
2
1 mol
in this reaction
10 g – 3.74 g = 6.3 g of copper will be leftover
4) What mass of silver will be produced?
n Ag(s): 0.118 mol x 2 = 0.118 mol x 107.87 g =
2
1 mol
Practice pg. 324 #4
12.7 g of silver
will be produced
in this reaction
 Putting it all together…
 In an experiment, 26.8g of iron (III) chloride in solution is combined
with 21.5g of sodium hydroxide. Which reactant is in excess, and by
how much? What mass of precipitate will be obtained?
FeCl3(aq) + 3NaOH(aq)  Fe(OH)3(s) + 3NaCl(aq)
26.8g
21.5g
m=?
162.20g/mol 40.00g/mol 106.88g/mol
nFeCl3 = 26.8g x 1 mol = 0.165 mol x 3 = 0.496 mol
162.20g
1
nNaOH = 21.5 g x 1 mol = 0.538 mol
40.0g
There is more
NaOH than
this so FeCl3
is the LR
0.538mol – 0.496 mol = 0.042 mol x 40.00 g = 1.7 g NaOH (excess)
1 mol
0.165 mol x 1 x 106.88 g
1
mol
= 17.7 g of Fe(OH)3(s)
 Identify the limiting reagent by choosing either
reagent amount, and use the mole ratio to compare the
required amount with the amount actually present.
 The quantity in excess is the difference between the
amount of excess reagent present and the amount
required for complete reaction.
 A reasonable reagent excess to use to ensure complete
reaction is 10%.
 Pg. 327 #4, 6-8