 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.
 When you know the quantity of more than one reagent,
you may need to know which one will limit the reaction.
1) 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
Use one of the amounts of reactants to test which reactant is
the limiting reagent
CH4
2 O2
=
x = 20 .0 mol
10.0 mol CH4
x mol O2
20.0 mol of oxygen
would be required
to react with 10.0
mol of methane
If we had used the
given amount of O2
to calculate the
amount of CH4
required, we would
find that 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? 10.0 mol – 5.00 mol = 5.00 mol
2) a) 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)
+ 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq)
10.0g
63.55 g/mol
n Cu(s):
20.0g
169.88g/mol
10.0g
= 0.157 mol
63.55 g/mol
n AgNO3(aq):
20.0g
= 0.118 mol
169.88 g/mol
That much silver
nitrate is not
available so
copper is not the
limiting reagent
More copper than
that is available so
silver nitrate is the
limiting reagent
x = 0.314 mol of AgNO3 are needed to react with 0.157 mol of Cu. We only
have 0.118 mol of AgNO3, therefore AgNO3 is the LIMITING REAGENT
b) 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)
XS
Cu(s)
0.157 mol
LR
+ 2 AgNO3(aq)  2 Ag(s) + Cu(NO3)2(aq)
0.118 mol
0.157 mol – 0.059 = 0.098 mol of Cu will be left over
m = 0.098 mol (63.55 g/mol) = 6.2g
c) What mass of silver will be produced?
Since AgNO3 is the limiting reactant, use it’s number of moles to determine how
much product (Ag) you can get.
x = 0.118 mol of Ag can be produced
m = n MM
= 0.118 mol (107.87 g/mol) = 12.7 g
 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
162.20g/mol
nNaOH = 21.5 g
40.0g/mol
= 0.165 mol
= 0.538 mol
Test one of the reactants to find out which one is the limiting reactant.
Therefore, 0.495 mol of NaOH are
Needed to react with 0.165 mol of
FeCl3. We have 0.538 mol of NaOH,
Therefore NaOH is the XS reagent and
FeCl3 is the LIMITING REAGENT.
Which reactant is in excess, and by how much?
NaOH is in excess.
Subtract the amount of NaOH given from the amount needed to react
with the number of moles of limiting reactant.
XS NaOH = 0.538mol NaOH – 0.496 mol NaOH = 0.042 mol NaOH
Find the mass of xs NaOH
m=n MM = 0.042 mol x 40.00 g/mol = 1.7 g NaOH (excess)
What mass of precipitate will be obtained?
Compare the number of moles of LR to
the amount of precipitate you can form
from the balanced equation.
Find the mass of precipitate (product)
m = n MM
= 0.165 mol x 106.88 g/mol
= 17.7 g of Fe(OH)3(s)
Balance Equation
2. Determine LR and XS reactant by testing the amount
of one of the products that the given amount of each
of the reactants can produce (REMEMBER THAT
ONLY MOLES CAN BE COMPARED). The LR is the
one that will produce less of the product.
3. Once you know the LR
1.
Find how much XS reactant there is by determining
how much of the XS reactant is needed to react with
the given amount of LR, and subtracting that amount
(in moles) from the given amount of XS reactant
B. Find how much product you can produce by using the
number of moles of LR in you molar ratio
A.
 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.
 Read 7.3 and 7.4
 Pg 330 # 2,3,6,7
 Pg 335 # 1,2,3