Limiting reagents

limiting reagent

1. Limiting Reagents
Caution: this stuff is difficult to follow at first.
Be patient.

2. g of
NaHCO
3
mL of 3M
HCl
1 10 25
2 10 50
3 10 100
How can we prove that our conclusions
about limiting reagents is correct?
Balloon & Flask Demonstration

3. Limiting reagent definedLimiting reagent defined
Q - How many moles of NO are produced if
__ mol NH
3
are burned in __ mol O
2
?
4 mol NH
3
, 5 mol O
2
4 mol NH
3
, 20 mol O
2
8 mol NH
3
, 20 mol O
2
Given: 4NH
3
+ 5O
2
® 6H
2
O + 4NO
4 mol NO, works out exactly
4 mol NO, with leftover O
2
8 mol NO, with leftover O
2
•Here, NH
3
limits the production of NO; if there was
more NH
3
, more NO would be produced
•Thus, NH
3
is called the “limiting reagent”
4 mol NH
3, 2.5 mol O
2
•In limiting reagent questions we use the limiting
reagent as the “given quantity” and ignore the
reagent that is in excess …
2 mol NO, leftover NH
3

4. Limiting reagents in stoichiometryLimiting reagents in stoichiometry
E.g. How many grams of NO are produced if
4 moles NH
3
are burned in 20 mol O
2
?
Since NH
3
is the limiting reagent we will use this
as our “given quantity” in the calculation
4NH
3
+ 5O
2
® 6H
2
O + 4NO
4 mol NO
4 mol NH
3
x
# g NO=
4 mol NH
3 = 120 g NO
30.0 g NO
1 mol NO
x
•Sometimes the question is more complicated.
For example, if grams of the two reactants are
given instead of moles we must first determine
moles, then decide which is limiting …

5. Solving Limiting reagents 1: g to molSolving Limiting reagents 1: g to mol
Q - How many g NO are produced if 20 g NH
3
is
burned in 30 g O
2
?
A - First we need to calculate the number of
moles of each reactant
4NH
3
+ 5O
2
® 6H
2
O + 4NO
1 mol NH
3

17.0 g NH
3
x # mol NH
3
=20 g NH
3

1.176
mol NH
3
=
1 mol O
2

32.0 g O
2
x # mol O
2
=30 g O
2
0.9375
mol O
2
=
A – Once the number of moles of each is
calculated we can determine the limiting
reagent via a chart …

6. NH
3
O
2
What we have
What we need
1.176 0.937
1.176/0.937
= 1.25 mol
0.937/0.937
= 1 mol
*Choose the smallest value to divide each by
** You should have “1 mol” in the same column
twice in order to make a comparison
4 5
4/5 = 0.8 mol 5/5 = 1 mol
A - There is more NH
3
(what we have) than
needed (what we need). Thus NH
3
is in
excess, and O
2
is the limiting reagent.
2: Comparison chart2: Comparison chart

7. 3: Stoichiometry (given = limiting)3: Stoichiometry (given = limiting)
So far we have followed two steps …
1) Expressed all chemical quantities as moles
2) Determined the limiting reagent via a chart
Finally we need to …
3) Perform the stoichiometry using the limiting
reagent as the “given” quantity
Q - How many g NO are produced if 20 g NH
3

is burned in 30 g O
2
?
4NH
3
+ 5O
2
® 6H
2
O + 4NO
4 mol NO
5 mol O
2

x
# g NO=
30 g O
2
22.5 g NO
=
30.0 g NO
1 mol NO
x
1 mol O
2

32.0 g O
2

x

8. Limiting Reagents: shortcutLimiting Reagents: shortcut
•Limiting reagent problems can be solved
another way (without using a chart)…
•Do two separate calculations using both given
quantities. The smaller answer is correct.
Q - How many g NO are produced if 20 g NH
3
is
burned in 30 g O
2
? 4NH
3
+ 5O
2
® 6H
2
O+ 4NO
4 mol NO
5 mol O
2

x 30 g O
2
22.5 g NO
=
30.0 g NO
1 mol NO
x
1 mol O
2

32.0 g O
2

x
4 mol NO
4 mol NH
3

x
# g NO=
20 g NH
3
35.3 g NO
=
30.0 g NO
1 mol NO
x
1 mol NH
3

17.0 g NH
3

x

9. Practice questionsPractice questions
1.2Al + 6HCl ® 2AlCl
3 + 3H
2
If 25 g of aluminum was added to 90 g of HCl, what
mass of H
2
will be produced (try this two ways –
with a chart & using the shortcut)?
2.N
2 + 3H
2 ® 2NH
3: If you have 20 g of N
2 and 5.0 g
of H
2, which is the limiting reagent?
3.What mass of aluminum oxide is formed when 10.0
g of Al is burned in 20.0 g of O
2?
4.When C
3H
8 burns in oxygen, CO
2 and H
2O are
produced. If 15.0 g of C
3
H
8
reacts with 60.0 g of
O
2
, how much CO
2
is produced?
5.How can you tell if a question is a limiting reagent
question vs. typical stoichiometry?

10. 11 1 mol Al
27.0 g Al
x # mol Al =25 g Al
= 0.926 mol
# mol HCl

=90 g HCl
1 mol HCl
36.5 g HCl
x
= 2.466 mol
Al HCl
What we
have
What we
need
0.926 2.466
0.926/0.926
= 1 mol
2.466/0.926
= 2.7 mol
2 6
2/2 = 1 mol 6/2 = 3 mol
HCl is
limiting.
3 mol H
2
6 mol HCl
x
# g H
2
=
90 g HCl
2.0 g H
2
1 mol H
2
x
1 mol HCl
36.5 g HCl
x = 2.47 g H
2

11. Question 1: shortcutQuestion 1: shortcut
2Al + 6HCl ® 2AlCl
3
+ 3H
2
If 25 g aluminum was added to 90 g HCl, what
mass of H
2
will be produced?
3 mol H
2
2 mol Al
x # g H
2
=25 g Al = 2.78 g H
2

2.0 g H
2
1 mol H
2
x
1 mol Al
27.0 g Al
x
3 mol H
2
6 mol HCl
x # g H
2
=90 g HCl = 2.47 g H
2

2.0 g H
2
1 mol H
2
x
1 mol HCl
36.5 g HCl
x

12. Question 2: shortcutQuestion 2: shortcut
N
2
+ 3H
2
® 2NH
3
If you have 20 g of N
2
and 5.0 g of H
2
, which is the
limiting reagent?
2 mol NH
3
1 mol N
2

x
# g NH
3
=
20 g N
2 = 24.3 g H
2

17.0 g NH
3
1 mol NH
3
x
1 mol N
2
28.0 g N
2
x
2 mol NH
3
3 mol H
2
x
# g NH
3
=
5.0 g H
2 = 28.3 g H
2

17.0 g NH
3
1 mol NH
3
x
1 mol H
2
2.0 g H
2
x
N
2
is the limiting reagent

13. Question 3: shortcutQuestion 3: shortcut
4Al + 3O
2
® 2 Al
2
O
3
What mass of aluminum oxide is formed when
10.0 g of Al is burned in 20.0 g of O
2
?
2 mol Al
2
O
3
4 mol Al
x
# g Al
2
O
3
=
10.0 g Al = 18.9 g Al
2
O
3

102.0 g Al
2
O
3
1 mol H
2
x
1 mol Al
27.0 g Al
x
2 mol Al
2
O
3
3 mol O
2
x
# g Al
2
O
3
=
20.0 g O
2 = 42.5 g Al
2
O
3
102.0 g Al
2
O
3
1 mol H
2
x
1 mol O
2
32.0 g O
2
x

14. Question 4: shortcutQuestion 4: shortcut
C
3
H
8
+ 5O
2
® 3CO
2
+ 4H
2
O
When C
3
H
8
burns in oxygen, CO
2
and H
2
O are
produced. If 15.0 g of C
3
H
8
reacts with 60.0 g
of O
2
, how much CO
2
is produced?
3 mol CO
2
1 mol C
3
H
8

x
# g CO
2
=
15.0 g C
3
H
8 = 45.0 g CO
2

44.0 g CO
2
1 mol CO
2
x
1 mol C
3
H
8
44.0 g C
3
H
8
x
3 mol CO
2
5 mol O
2
x
# g CO
2
=
60.0 g O
2 = 49.5 g CO
2
44.0 g CO
2
1 mol CO
2
x
1 mol O
2
32.0 g O
2
x
5. Limiting reagent questions give values for
two or more reagents (not just one)

15. N
2
H
2
What we have
What we need
Question 2Question 2
0.714 mol 2.5 mol
0.714/0.714
= 1 mol
2.5/0.714
= 3.5 mol
We have more H
2
than what we need, thus H
2
is
in excess and N
2 is the limiting factor.
1 mol 3 mol
1 mol N
2

28 g N
2
x # mol N
2
=20 g N
2
0.714 mol N
2
=
1 mol H
2

2 g H
2
x # mol H
2
=5.0 g H
2 2.5 mol H
2
=

16. Al O
2
334Al + 3O
2
® 2 Al
2
O
3
1 mol Al
27 g Al
x # mol Al =10 g Al
0.37 mol Al
=
1 mol O
2

32 g O
2
x # mol O
2
=20 g O
2
0.625 mol O
2=
0.37 mol 0.625 mol
0.37/.37 = 1
mol
0.625/0.37
= 1.68 mol
4 mol 3 mol
4/4 = 1 mol3/4 = 0.75 mol
What we
have
What we
need
There is
more
than
enough
O
2
; Al is
limiting
2 mol Al
2
O
3
4 mol Al
x # g Al
2
O
3
=0.37 mol Al
18.87 g Al
2
O
3
=
102 g Al
2
O
3
1 mol Al
2
O
3
x

17. C
3
H
8
O
2
44 C
3
H
8
+ 5O
2
® 3CO
2
+ 4H
2
O
1 mol C
3
H
8

44 g C
3
H
8
x # mol C
3
H
8
=15 g C
3
H
8
0.34 mol
C
3
H
8
=
1 mol O
2

32 g O
2
x # mol O
2
=60 g O
2
1.875 mol O
2=
0.34 mol 1.875 mol
0.34/.34 = 1
mol
1.875/0.34 =
5.5 mol
1 mol 5 mol
What we
have
Need
3 mol CO
2
1 mol C
3
H
8

x
# g CO
2
=
0.34 mol C
3
H
8
45 g CO
2
=
44 g CO
2
1 mol CO
2
x
We have
more than
enough O
2
,
C
3
H
8
is
limiting

18. Limiting Reagents: shortcutLimiting Reagents: shortcut
MgCl
2
+ 2AgNO
3
® Mg(NO
3
)
2
+ 2AgCl
If 25 g magnesium chloride was added to 68 g
silver nitrate, what mass of AgCl will be
produced?
2 mol AgCl
1 mol MgCl
2

x
# g AgCl=
25 g MgCl
2
75.25 g AgCl
=
143.3 g AgCl
1 mol AgCl
x
1 mol MgCl
2
95.21 g MgCl
2

x
2 mol AgCl
2 mol AgNO
3

x
# g AgCl=
68 g AgNO
3
57.36 g AgCl
=
143.3 g AgCl
1 mol AgCl
x
1 mol AgNO
3
169.88 g AgNO
3
x
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