Stoichiometry
Chapter 11
Natural Approach to
Chemistry
Assignments:
11.1 362/38cd, 39cd, 43acde, 40cd
11.2 363/46-48; 363/49,50ab,51
11.3 364/59-62
11.4 365/64 and 66
1
2
Learning Objectives
• Apply the mole concept and the law of
conservation of mass to calculate quantities of
chemicals participating in reactions.
• Important terms: stoichiometry, percent yield,
actual yield, theoretical yield, limiting reactant
3
Chemical equations tell stories…
2CO(g) + O2(g) → 2CO2(g)
… and stories can be put into different categories
Nonfiction
Synthesis / Decomposition
Single / Double replacement
Precipitate reaction
Science fiction
Adventure
Romance
History
Psychology
Polymerization reaction
Children’s literature
…
4
Chemical equations tell stories…
But what exactly do they tell us?
2CO(g) + O2(g)
2CO2(g)
They tell us what compounds we start with:
Carbon monoxide (CO) gas
Oxygen (O2) gas
what compounds are formed:
Carbon dioxide (CO2) gas
5
Chemical equations tell stories…
What else do they tell us?
2CO(g) + O2(g)
2 CO molecules 1 O2 molecules
2CO2(g)
2 CO2 molecules
They tell us how much of each compound is involved
stoichiometry: the study of the amounts of
substances involved in a chemical reaction.
6
2CO(g) +
O2(g)
2 CO molecules
2 dozen CO molecules
2 moles CO molecules
2 x (6.023 x 1023) CO molecules
2CO2(g)
2 CO2 molecules
2 dozen CO2 molecules
2 moles CO2 molecules
2 x (6.023 x 1023) CO2 molecules
1 O2 molecules
1 dozen O2 molecules
1 mole O2 molecules
(1 x) 6.023 x 1023 O2 molecules
7
Is that
okay?
2CO(g) + O2(g)
Number of
moles is
not
conserved
2 moles
CO molecules
+
1 mole
O2 molecules
2CO2(g)
≠
2 moles
CO2 molecules
Yes, as long as the chemical equation is balanced!
The coefficients are important!!!
8
These are
important!
Coefficients
2CO(g) + O2(g)
2CO2(g)
2 moles
1 mole
2 moles
CO molecules
O2 molecules
CO2 molecules
This chemical equation is balanced
The coefficients are correct
9
Coefficients are important
1 bag
cake mix
+
3 eggs + ¼ cup oil
Write as a ratio:
+
1 cup water
1 cup oil
4
1 batch cupcakes
or
1 batch cupcakes
1 batch cupcakes
1 cup oil
4
10
Coefficients are important
1 bag
cake mix
+
3 eggs + ¼ cup oil
+
1 cup water
1 cup oil
Write as a ratio:
4 batches cupcakes
or
1 batch cupcakes
4 batches cupcakes
1 cup oil
11
Fermentation of sugar (glucose) into alcohol:
C6H12O6(aq)
1 mole
glucose
2C2H5OH(aq) + 2CO2(g)
2 moles
ethanol
Write as a ratio:
1 mole glucose
2 moles ethanol
2 moles
carbon
dioxide
2 moles ethanol
1 mole glucose
These are stoichiometric equivalents
12
Fermentation of sugar (glucose) into alcohol:
C6H12O6(aq)
1 mole
glucose
Write as a ratio:
2C2H5OH(aq) + 2CO2(g)
2 moles
ethanol
1 mole glucose
2 moles ethanol
2 moles
carbon dioxide
2 moles ethanol
1 mole glucose
mole ratio: a ratio comparison between substances
in a balanced equation. It is obtained from the
coefficients in the balanced equation.
13
Mole ratios
Fermentation of sugar (glucose) into alcohol:
C6H12O6(aq)
1 mole
glucose
mole
ratios for
this
chemical
equation
2C2H5OH(aq) + 2CO2(g)
2 moles
ethanol
1 mole glucose
2 moles ethanol
1 mole glucose
2 moles CO2
2 moles ethanol
2 moles CO2
2 moles
carbon dioxide
2 moles ethanol
1 mole glucose
2 moles CO2
1 mole glucose
2 moles CO2
2 moles ethanol
14
11.1 Analyzing a Chemical Reaction
Mole ratios
Consider the following equation:
CO(g) + 2H2(g)
carbon
monoxide
hydrogen
CH3OH(l)
methanol
If the reaction produces 5 moles of CH3OH, how many moles of
H2 are consumed?
Asked: moles of H2
5 moles CH3OH x 2 moles H2
1 mole CH3OH
=
10 moles H2
15
A mixture of aluminum metal and chlorine gas reacts to form
aluminum chloride (AlCl3): 2Al(s) + 3Cl2(g) → 2AlCl3(s). How many
moles of aluminum chloride will form when 5 moles of chlorine gas
react with excess aluminum metal?
Asked: moles AlCl3
Given: moles Cl2
5 mole Cl2 x 2 mole AlCl3 = 3.3 mole AlCl3
3 mole Cl2
16
Potassium + Hydrogen Phosphate 
Finish the reaction in symbols and balance…
If 14.72 moles of hydrogen phosphate are consumed in the
above reaction, how many moles of hydrogen are
produced?
17
18
mass  moles….
There is no scale that measures in moles!
How do you convert from moles to grams?
By using the molar mass (g/mole)
The mass of 1 mole of Al is not the same as the mass of 1
mole of Cl2.
How do you convert from grams of Al to grams of Cl2?
By using the molar mass (g/mole) and mole ratios
19
Process for calculating grams from grams given
20
If 45.0 g of calcium carbonate (CaCO3) decomposes in the reaction
CaCO3(s) → CaO(s) + CO2(g), how many grams of CO2 are
produced?
Asked: grams of CO2
Given: grams of CaCO3
Relationships:
mole ratios
molar mass of CaCO3 = 40.078 + 12.011 + (15.999 x 3) =
100.0 g/mole
molar mass of CO2 = 12.011 + (15.999 x 2) = 44.01 g/mole
Strategy:
21
If 45.0 g of calcium carbonate (CaCO3) decomposes in the reaction
CaCO3(s) → CaO(s) + CO2(g), how many grams of CO2 are
produced?
B
B
0.45 mole CaC03 x 1 mole CO2 x 44.01 g CO2 = 19.8 g CO2
1 mole CaC03
1 mole C02
22
CHAPTER 11
Stoichiometry
11.2 Percent Yield and
Concentration
23
In theory, all 100 kernels should have popped.
Did you do something wrong?
+
100 kernels
82 popped
18 unpopped
24
In theory, all 100 kernels should have popped.
Did you do something wrong?
No
In real life (and in the lab)
things are often not perfect
+
100 kernels
82 popped
18 unpopped
25
Percent yield
What you get to eat!
percent yield 
amount of corn popped
 100
amount of kernels in the bag
82
percent yield 
 100  82%
100
+
100 kernels
82 popped
18 unpopped
26
actual yield
percent yield 
 100
theoretical yield
percent yield 
actual yield
 100
theoretical yield
actual yield: the amount obtained in the lab in an
actual experiment.
theoretical yield: the expected amount produced if
everything reacted completely.
27
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
Heating
28
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
4.87 g
10.00 g
measured experimentally
Can you think of reasons why the final mass of Na2CO3 may
not be accurate? (What could be sources of error?)
-There is usually some human error, like not measuring exact amounts
- carefully
-Maybe the heating time was not long enough; not all the Na2HCO3
- reacted
- Maybe Na2CO3 was not completely dry; some H2O(l) was measured too
- CO2 is a gas and does not get measured
29
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
4.87 g
measured experimentally
Let’s calculate the percent yield
obtained in experiment
actual yield
percent yield 
 100
theoretical yield
calculated
30
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
4.87 g
measured experimentally
Let’s calculate the percent yield
4.87 g
percent yield 
 100
theoretical yield
calculated
31
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s)
→ Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
4.87 g
measured experimentally
Let’s calculate the percent yield
4.87 g
percent yield 
 100
theoretical yield
calculated
32
10.00 g
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
This is a gram-to-gram conversion:
molar mass of NaHCO3  22.99  1.0079  12.011  3(15.999)  84.01 g / mole
molar mass of Na2CO
22.99  2   12.011  15.999  3   105.99 g / mole
3  NaHCO
1 mole
3
10.00 g NaHCO3 
 0.1190 moles NaHCO3
84.01 g105.99
NaHCO
g 3Na2CO3
0.05950 moles Na2CO3 
 6.306 g Na2CO3
1 mole Na2CO3
10.00g NaHCO3 1 mole NaHCO3
84.01 g NaHCO3
1 mole Na2CO3 1mole Na2CO3 =
2 mole NaHCO3 105.99 g Na2CO3
Answer: Mass of D = 6.306 g Na2CO3
33
2NaHCO3(s)
→ Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
10.00 g
0.1190 moles
4.87 g
0.05950 moles
6.306 g
For 10.00 g of starting material (NaHCO3), the
theoretical yield for Na2CO3 is 6.306 g.
The actual yield (measured) is 4.87 g.
34
2NaHCO3(s)
10.00 g
→ Na2CO3(s) + H2O(l) + CO2(g)
4.87 g
actual yield
percent yield 
 100
theoretical yield
4.87 g
percent yield 
 100  77.2%
6.306 g
For 10.00 g of starting material (NaHCO3), the
theoretical yield for Na2CO3 is 6.306 g.
The actual yield (measured) is 4.87 g.
35
Stoichiometry with solutions
Decomposition of baking soda: (We just looked at this.)
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
Convert to moles
Reaction of solid zinc with hydrochloric acid:
Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq)
50.0 mL of a
3.0 M
solution
Reactions in
solution
Convert to moles
36
A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0
M solution of hydrochloric acid (HCl) according to:
Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq). How many grams of
hydrogen gas (H2) will be produced? Assume zinc metal is
present in excess.
Asked: grams of H2 produced
Given: 50.0 mL of 3.0 M HCl
Reacting with excess zinc
Solve:
Relationships:
M = mole/L
Mole ratio: 2 moles HCl ~ 1 mole H2
Molar mass of H2 = 1.0079 x 2
= 2.02 g/mole
0.0500L HCl x 3.0mole HCl x 1 mol H2 x 2.02 g H2 = 0.15 g H2
L HCl
2 mol HCl
1 mol H2
Answer: 0.15 grams of H2 are produced
37
38
Reaction of solid zinc with hydrochloric acid:
Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq)
50.0 mL of a
3.0 M
Convert molarity to moles
solution
Sometimes the concentration is written in mass percent
mass of compound
mass % of compound 
 100
total mass of solution
Vinegar is 5% acetic acid by mass
39
Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by
mass. How many grams of acetic acid are in 120 mL of commercial
vinegar? (Assume the density of vinegar is the same as pure water,
1.0 g/mL.)
Asked: grams of acetic acid in 120 mL of vinegar
Given: 120 mL of vinegar and 5% acetic acid by mass
mass of acetic acid
Relationships: 120 mL = 120 g,
mass % 
 100
mass of solution
given a density of 1.0 g/mL
Solve: mass %  mass of acetic acid
100
mass of solution
5 % mass of acetic acid

100
120 g
0.05 
mass of acetic acid
120 g
mass of acetic acid  0.05  120 g
mass of acetic acid  6.0 g
Answer: 6.0 g of acetic acid.
40
Let’s Review:
percent yield 
actual yield
 100
theoretical yield
Obtained from the
experiment
actual yield
percent yield 
 100
theoretical yield
Calculate using molar
masses and mole ratios
mass of compound
mass % of compound 
 100
total mass of solution
41
Assignments 11.2:
42
Limiting Reactants
Ch 11.3
Suppose you want to make 2 ham & cheese sandwiches
Can you still make 2 ham & cheese sandwiches if you
have 4 slices of bread, 4 slices of ham, and 1 slice of
cheese?
Limiting factor
No, you are limited by the cheese!
You can only get 1 ham & cheese sandwich.
Limiting
reactant
Excess
reactant
limiting reactant: the reactant that “runs
out” first in a chemical reaction.
excess reactant: the reactant that is
remaining after the reaction is complete.
Steps for Determining the Limiting Reactant
Step 1
Step 2
Step 3
Convert both
reactant masses to
moles.
Multiply by the mole
ratio from the
balanced equation
to find how much
reactant is needed
to use up all of the
other reactant.
Compare the
amounts of
reactants. Compare
what you have
available to what
you need.
This gives you the
amount you have
available to use.
This gives you the
amount you need to
consume all of the
reactant.
If what you need is
more than what you
have, then this is the
limiting reactant.
Sample Problem: 364/58.
Iron can be produced from the following reaction:
Fe2O3 + 2Al  2Fe + Al2O3
a. If 100.0 g of Fe2O3 reacts with 30.0 g Al, which one
will be used up first?
Sample Problem: 365/58.
Iron can be produced from the following reaction:
Fe2O3 + 2Al  2Fe + Al2O3
b. For this next step, use the smaller mole answer from
a. to find the needed amount
c. How much Fe can be produced?
1.112 mole Al
2 mole Fe
2 mole Al
55.85 g Fe
1 mole Fe
= 62.10 g Fe
d. How much of the excessive reactant is remaining?
Fe2O3 is the excessive reactant.
mole Al  mole Fe2O3  mass Fe2O3
Have – used = excess
1.112 mole Al 1 mole Fe2O3 159.70 g Fe2O3 = 88.795gFe2O3
2 mole Al
1 mole Fe2O3
100.0g – 88.80 g = 11.20 g remaining (excess)
Asgn: 364/59, 60
11.3 Assignment
361/31,34,58-62
Solving
Stoichiometric
Problems
CHAPTER 14.4
Section 11.1
Analyzing a Chemical Reaction
Section 11.2
Percent Yield and Concentration
Section 11.3
Limiting Reactants
Section 11.4
Solving Stoichiometric Problems
Use what we’ve learned to answer
these questions:
- What is the limiting reactant?
- What is the theoretical yield?
- What is the percent yield?
- How much excess reactant is left?
- How much reactant is used if it’s in a
solution?
Section 11.1
Analyzing a Chemical Reaction
Section 11.2
Percent Yield and Concentration
Section 11.3
Limiting Reactants
Section 11.4
Solving Stoichiometric Problems
Use what we’ve learned to answer
these questions:
- What is the limiting reactant?
- What is the theoretical yield?
- What is the percent yield?
- How much excess reactant is left?
- How much reactant is used if it’s in a
solution?
What is the limiting reactant?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce
lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the
limiting reactant? Lithium is the only group 1 metal that is capable of
reacting directly with nitrogen gas.
Solve:
1) Moles of each reactant?
2) Apply the mole ratio
3) Compare what we have
with what we need
What is the limiting reactant?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce
lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the
limiting reactant? Lithium is the only group 1 metal that is capable of
reacting directly with nitrogen gas.
Asked: Limiting reactant
Given: 48.0 g of Li (have)
46.5 g of N2 (have)
molar mass of Li = 6.941 g/mole
molar mass of N2 = 28.01 g/mole
mole ratio: 6 moles Li ~ 1 mole N2
What is the limiting reactant?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce
lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
48.0 g of Li reacts with 46.5 g of N2
1 mole
48.0 g 
 6.92 moles Li (have )
6.941 g
46.5 g 
1 mole
 1.66 moles N2 (have )
28.01 g
What is the limiting reactant?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to
produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
? We have: 6.92 moles Li; 1.66 moles N
2
2) Apply the mole ratio
How much N2 do we need to react with 6.92 moles Li?
1 mole N2
6.92 moles Li 
 1.15 moles N2 (need )
6 moles Li
Do we have enough N2?
Yes, we have more than enough N2. That means we will run
out of Li before we run out of N2
Section 11.1
Analyzing a Chemical Reaction
Section 11.2
Percent Yield and Concentration
Section 11.3
Limiting Reactants
Section 11.4
Solving Stoichiometric Problems
Use what we’ve learned to answer
these questions:
- What is the limiting reactant?
- What is the theoretical yield?
- What is the percent yield?
- How much excess reactant is left?
- How much reactant is used if it’s in a
solution?
What is the theoretical yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to
produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
How much lithium nitride (LiN3) can be produced from this
reaction?
What is the theoretical yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N)
according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
How much lithium nitride (Li3N) can be produced from this reaction?
Asked:
Amount of Li3N produced
Given:
Li is the limiting reactant
6.92 moles Li (have)
molar mass of Li 3N
From
the last
problem
Relationships:
molar mass of Li3N = 34.83 g/mole
mole ratio: 6 moles Li ~ 2 moles Li3N
g 
g

 3   6.941

14.007
mole 
mole

g
 34.83
mole
What is the theoretical yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N)
according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
How much lithium nitride (Li3N) can be produced from this reaction?
Asked:
Amount of Li3N produced
Given:
Li is the limiting reactant
6.92 moles Li (have)
Relationships:
molar mass of Li3N = 34.83 g/mole
mole ratio: 6 moles Li ~ 2 moles Li3N
Solve:
1) Find moles of Li3N
2) Convert moles to grams
What is the theoretical yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N)
according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked:
Amount of Li3N produced
Given:
Li is the limiting reactant
6.92 moles Li (have)
6.92 moles Li 
2.31 moles 
Relationships:
molar mass of Li3N = 34.83 g/mole
mole ratio: 6 moles Li ~ 2 moles Li3N
Solve:
1) Find moles of Li3N
2) Convert moles to grams
2 moles Li 3N
 2.31 moles Li 3N
6 moles Li
34.83 g
 80.46 g Li 3N
1 mole
What is the theoretical yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N)
according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked:
Amount of Li3N produced
Given:
Li is the limiting reactant
6.92 moles Li (have)
6.92 moles Li 
2.31 moles 
Relationships:
molar mass of Li3N = 34.83 g/mole
mole ratio: 6 moles Li ~ 2 moles Li3N
Solve:
1) Find moles of Li3N
2) Convert moles to grams
2 moles Li 3N
 2.31 moles Li 3N
6 moles Li
34.83 g
 80.46 g Li 3N
1 mole
What is the theoretical yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N)
according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked:
Amount of Li3N produced
Given:
Li is the limiting reactant
6.92 moles Li (have)
6.92 moles Li 
2.31 moles 
Relationships:
molar mass of Li3N = 34.83 g/mole
mole ratio: 6 moles Li ~ 2 moles Li3N
Solve:
1) Find moles of Li3N
2) Convert moles to grams
Asked:
2 moles Li 3N
 2.31 moles Li 3N
6 moles Li
34.83 g
 80.46 g Li 3N
1 mole
80.46 g of Li3N are produced
Section 11.1
Analyzing a Chemical Reaction
Section 11.2
Percent Yield and Concentration
Section 11.3
Limiting Reactants
Section 11.4
Solving Stoichiometric Problems
Use what we’ve learned to answer
these questions:
- What is the limiting reactant?
- What is the theoretical yield?
- What is the percent yield?
- How much excess reactant is left?
- How much reactant is used if it’s in a
solution?
What is the percent yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N)
according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Calculate the percent yield of an experiment that actually produced 62.5 g of Li3N.
What is the percent yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N)
according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Calculate the percent yield of an experiment that actually produced 62.5 g of Li3N.
Asked:
Percent yield
Given:
Theoretical yield: 80.46 g Li3N
Actual yield: 62.5 g Li3N
From the
last problem
Relationships:
percent yield 
Solve:
actual yield
 100
theoretical yield
Use the percent yield formula
What is the percent yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N)
according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked:
Percent yield
Given:
Theoretical yield: 80.46 g Li3N
Actual yield: 62.5 g Li3N
Relationships:
percent yield 
Solve:
actual yield
 100
theoretical yield
Use the percent yield formula
percent yield 
62.5 g (actual )
 100  77.7%
80.46 g (theoretical )
What is the percent yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N)
according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked:
Percent yield
Given:
Theoretical yield: 80.46 g Li3N
Actual yield: 62.5 g Li3N
Answer: The percent yield in this
particular experiment is 77.7%
Relationships:
percent yield 
Solve:
actual yield
 100
theoretical yield
Use the percent yield formula
percent yield 
62.5 g (actual )
 100  77.7%
80.46 g (theoretical )
Section 11.1
Analyzing a Chemical Reaction
Section 11.2
Percent Yield and Concentration
Section 11.3
Limiting Reactants
Section 11.4
Solving Stoichiometric Problems
Use what we’ve learned to answer
these questions:
- What is the limiting reactant?
- What is the theoretical yield?
- What is the percent yield?
- How much excess reactant is left?
- How much reactant is used if it’s in a
solution?
What is left?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to
produce lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
How much of the excess reactant remains after the limiting
reactant is completely consumed?
What is left?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce
lithium nitride (Li3N) according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
How much of the excess reactant remains after the limiting reactant is
completely consumed?
Asked: Amount of excess
Solve: 1) How many moles N2 remain?
reactant left
2) Convert moles to grams
Given: N2 is the excess reactant
1.15 moles N2 (need)
1.66 moles N2 (have)
Relationships:
Molar mass of N2 = 28.01 g/mole
What is left?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N)
according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked:
Given:
Amount of excess reactant left
1.66 moles N2 (have)  1.15 moles
N2 is the excess reactant
1.15 moles N2 (need)
1.66 moles N2 (have)  0.51 moles N2 ( remaining )
Relationships:
28.01 g/mole N2
Solve:
1) How many moles N2 remain?
2) Convert moles to grams
N2 (need )
What is left?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N)
according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked:
Amount of excess reactant left
Given:
N2 is the excess reactant
1.15 moles N2 (need)
1.66 moles N2 (have)
Relationships:
28.01 g/mole N2
Solve:
1.66 moles N2 (have )  1.15 moles N2 (need )
 0.51 moles N2 (remaining )
28.01 g
0.51 moles N2 
 14 g N2
1 mole
1) How many moles N2 remain?
2) Convert moles to grams
What is left?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N)
according to the reaction:
6Li(s) + N2(g) → 2Li3N(s)
Asked:
Amount of excess reactant left
Given:
N2 is the excess reactant
1.15 moles N2 (need)
1.66 moles N2 (have)
Relationships:
28.01 g/mole N2
Solve:
1.66 moles N2 (have )  1.15 moles N2 (need )
 0.51 moles N2 (remaining )
0.51 moles N2 
28.01 g
 14 g N2
1 mole
1) How many moles N2 remain?
2) Convert moles to grams
Answer:
14 g of N2 will remain at the end of the reaction.
365/64. Aluminum metal reacts with oxygen in the air to form a layer
of aluminum oxide according to the equation:
4Al + 3O2 --> 2Al203(s)
A. If 28.0 g of Al reacts with excess oxygen in the air, what mass of
aluminum oxide is formed?
B. Molar Mass Al203 = 2Al+3(0) = 2(27)+3(16) = 102 g/mole
28.0gAl 1mole Al 2 mole Al203
27g Al
4 mole Al
102 g Al203 = 52.9 g Al203
1 mole Al203
365/64. Aluminum metal reacts with oxygen in the air to form a
layer of aluminum oxide according to the equation:
4Al + 3O2 --> 2Al203(s)
B How many moles of oxygen are consumed
during this reaction?
28 g Al 1mole Al 3mole O2 = 0.778 mole O2
27 g Al 4 mole Al
365/64. Aluminum metal reacts with oxygen in the air to form a layer
of aluminum oxide according to the equation:
4Al + 3O2 --> 2Al203(s)
C. If 250 g Al203 of is formed, how much Al
reacted?
250 g Al203 1 mole Al203 4 mole Al
27.0 g Al
102 g Al203 2 mole Al203 1 mole Al
Answer: 132.4 g Al
365/66. Wine can spoil when ethanol is converted to acetic
acid by oxidation:
C2H5OH + O2 --> CH3COOH + H2O
a. Determine the limiting reactant when 5.00 g of
ethanol (C2H5OH) and 2.0 g of oxygen are sealed
in a bottle.
2C + 6H + O = 24 + 6 + 16 = 46.1 g/mole
5.00g eth. 1 mole eth. = 0.108 moles ethanol
46.1 g eth
2.0 g oxygen 1 mole oxy. = 0.0625 moles oxygen
32.0g oxy.
O2 is the limiting reactant.
We converted reactants to moles individually.
365/66. Wine can spoil when ethanol is converted to acetic acid
by oxidation:
C2H5OH + O2 --> CH3COOH + H2O
b. Calculate how much acetic acid will form in grams. Start with the
oxygen because it is the limiting reactant.
0.0625 moles O2 1 mole CH3COOH 60 g CH3COOH
1 mole O2
1 mole CH3COOH
Answer: 3.75 g
Molar mass: 2C + 4H + 2(O) =24+4+32 = 60 g/mole
365/66. Wine can spoil when ethanol is converted to acetic acid by
oxidation:
C2H5OH + O2 --> CH3COOH + H2O
c. Calculate the amount of excess reactant
remaining in grams.
Excess reactant is ethanol.
Available less used:
0.109 moles – 0.0625 moles = 0.0465 moles
0.0465 moles eth. 46 g eth.
= 2.139 g eth.
1 mole eth.
Amount of excess ethanol is 2.139 g
Section 11.1
Analyzing a Chemical Reaction
Section 11.2
Percent Yield and Concentration
Section 11.3
Limiting Reactants
Section 11.4
Solving Stoichiometric Problems
Use what we’ve learned to answer
these questions:
- What is the limiting reactant?
- What is the theoretical yield?
- What is the percent yield?
- How much excess reactant is left?
- How much reactant is used if it’s in a
solution?
Reactants in solution
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from
an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to
1.0 L of the contaminated water. What is the concentration of CuSO4 in the water if 0.021 g
of CuS precipitate is formed? The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Reactants in solution
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water
discharged from
an industrial plant is found by adding an excess of sodium sulfide
(Na2S) solution to
1.0 L of the contaminated water. What is the concentration of CuSO4
in the water if 0.021 g of CuS precipitate is formed? The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Asked:
Concentration of CuSO4(aq)
molar mass of CuS
Given:
1.0 L of solution is tested
0.021 g CuS (formed)
 63.55  32.06  95.61
Relationships:
Molar mass of CuS = 95.61 g/mole
Mole ratio: 1 mole CuSO4 ~ 1 mole CuS
g
mole
Reactants in solution
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from
an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to
1.0 L of the contaminated water. What is the concentration of CuSO4 in the water if 0.021 g
of CuS precipitate is formed? The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Asked:
Concentration of CuSO4(aq)
Given:
1.0 L of solution is tested
0.021 g CuS (formed)
Solve:
Relationships:
Molar mass of CuS = 95.61 g/mole
Mole ratio: 1 mole CuSO4 ~ 1 mole CuS
1) How many moles of CuS?
2) How many moles of CuSO4?
3) What is the concentration of
CuSO4?
Reactants in solution
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an
industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the
contaminated water. The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Asked:
Concentration of CuSO4(aq)
Given:
1.0 L of solution is tested
0.021 g CuS (formed)
Relationships:
95.61 g/mole CuS
1 mole CuSO4 ~ 1 mole CuS
Solve:
CuSO4?
1) How many moles of CuS?
2) How many moles of CuSO4?
3) What is the concentration of
0.021 g CuS 
1 mole
 2.20  104 moles CuS
95.61 g
Reactants in solution
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an
industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the
contaminated water. The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Asked:
Concentration of CuSO4(aq)
Have: 2.20  104 moles CuS
Given:
1.0 L of solution is tested
0.021 g CuS (formed)
2.20  104 moles CuS 
Relationships:
95.61 g/mole CuS
1 mole CuSO4 ~ 1 mole CuS
Solve:
CuSO4?
1) How many moles of CuS?
2) How many moles of CuSO4?
3) What is the concentration of
1 mole CuSO4
1 mole CuS
 2.20  104 moles CuSO4
Reactants in solution
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an
industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the
contaminated water. The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Asked:
Concentration of CuSO4(aq)
Given:
1.0 L of solution is tested
0.021 g CuS (formed)
Relationships:
95.61 g/mole CuS
1 mole CuSO4 ~ 1 mole CuS
Solve:
CuSO4?
1) How many moles of CuS?
2) How many moles of CuSO4?
3) What is the concentration of
Have: 2.20  104 moles CuS
2.20  104 moles CuSO4
molarity 
moles of solute
L of solution
2.20  104 moles CuSO4

1.0 L of water
 2.20  104 M CuSO4
Reactants in solution
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an
industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the
contaminated water. The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Asked:
Concentration of CuSO4(aq)
Given:
1.0 L of solution is tested
0.021 g CuS (formed)
Relationships:
95.61 g/mole CuS
1 mole CuSO4 ~ 1 mole CuS
Solve:
CuSO4?
1) How many moles of CuS?
2) How many moles of CuSO4?
3) What is the concentration of
Answer: The concentration of CuSO4
2.20 x 10-4 M.
molarity 
moles of solute
L of solution
2.20  104 moles CuSO4

1.0 L of water
 2.20  104 M CuSO4
is
Reactants in solution
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an
industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the
contaminated water. The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Discussion:
If the legal limit is 5.0 x 10-4 M of Cu2+, is
the industrial plant following the
environmental guidelines?
Answer: The concentration of CuSO4
2.20 x 10-4 M.
is
Reactants in solution
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an
industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the
contaminated water. The reaction is:
CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
Discussion:
If the legal limit is 5.0 x 10-4 M of Cu2+, is
the industrial plant following the
environmental guidelines?
Answer: The concentration of CuSO4
2.20 x 10-4 M.
Yes, because 2.20 x 10-4 M is less than the
legal limit.
CuSO4(aq) → Cu2+(aq) + SO42–(aq)
2.20  104 M
2.20  104 M
is