Chapter 5:
Chemical
Reactions and
Equations
5-1
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Questions for Consideration
1.
2.
3.
4.
5.
What happens in a chemical reaction?
How do we know whether a chemical
reaction takes place?
How do we represent a chemical reaction
with a chemical equation?
How are chemical reactions classified?
How can the products of different classes
of chemical reactions be predicted?
How do we represent chemical reactions in
aqueous solution?
5-2
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Chapter 5 Topics:
1.
2.
3.
4.
5.
What is a Chemical Reaction?
How Do We Know a Chemical Reaction
Occurs?
Writing Chemical Equations
Predicting Chemical Reactions
Representing Reactions in Aqueous
Solution
5-3
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Introduction
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Chemical reactions
occur all around us.
How do we make sense
of these changes?
What patterns can we
find?
Figure 5.1
Figure 5.29
Figure 5.8F
5-4
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5.1 What is a Chemical Reaction?
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A chemical reaction is a chemical change.
A chemical reaction occurs when one or
more substances is converted into one or
more new substances.
Reactant – a substance that we start with
that undergoes a change
Product – a new substance that forms
during the reaction
Products differ from reactants only in the
arrangement of their component atoms.
5-5
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5.2 How do We Know a Chemical
Reaction Occurs?

What observations might indicate that a chemical
reaction has taken place?
Figure 5.7
5-6
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What clues tell you a reaction is
likely occurring?
Figure 5.8
5-7
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Physical Clues of a Chemical Reaction

Common observations that may accompany a
chemical reaction are:
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change in color
production of light
formation of a solid (such as a precipitate in
solution, or smoke in air, or a metal coating)
formation of a gas (bubbles in solution or fumes in
the gaseous state)
absorption or release of heat (sometimes
appearing as a flame)
5-8
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5.3 Writing Chemical Equations

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A chemical equation is a symbolic
representation of a chemical reaction.
A chemical equation shows:
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the formulas for reactants and products
the physical states of each substance (s, l, g, aq)
relative numbers of reactants that combine and
products that form
special conditions required for the reaction,
such as constant heating.
5-9
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Chemical Reactions


When hydrogen gas is ignited in the presence of
oxygen, an explosive reaction occurs producing
gaseous water molecules.
Is mass conserved?
Replace with updated Figure 5.6
Figure 5.5
Figure 5.6
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5-10
Writing Chemical Equations

Balanced equation
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The number of atoms of each element is the
same in the products as in the reactants.
Conservation of mass is always maintained.
5-11
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Writing Chemical Equations

When a powdered mixture of
aluminum metal and
iron(III) oxide is heated, it
reacts to form liquid iron
metal and aluminum(III)
oxide.
Al(s) + Fe2O3(s)  Al2O3(s) + Fe(l)
Figure 5.9
5-12
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Chemical Equations Must be Balanced

Al(s) + Fe2O3(s)  Al2O3(s) + Fe(l)
Figure from p. 177


Is this equation balanced?
What has to change?
5-13
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Balance Equations with Coefficients
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Al(s) + Fe2O3(s)  Al2O3(s) + Fe(l)
We balance the atoms in equations with
coefficients.

The aluminum atoms are not balanced so we place a
coefficient of 2 in front of the Al reactant:
2Al(s) + Fe2O3(s)  Al2O3(s) + Fe(l)

The iron atoms are not balanced so we place a
coefficient of 2 in front of the Fe product:
2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(l)
Figure
from
p. 177
5-14
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Writing Chemical Equations
Methane + oxygen  carbon dioxide + water
CH4(g) + O2(g)  CO2(g) + H2O(g)
The number of atoms of each element in the unbalanced
equation is shown below. These numbers were obtained by
multiplying the subscript to the right of the element’s
symbol by the stoichiometric coefficient.
# of atoms (reactants)
1C
4H
# of atoms (products)
1C
2H
2O
3O
5-15
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Writing Chemical Equations
CH4(g) + O2(g)  CO2(g) + H2O(g)
First, we look at the carbon atoms. Since the number of
carbon atoms on the reactant side is already equal to the
number of carbon atoms on the product side, we don’t
need to add coefficients.
Next, we look at the hydrogen atoms. Currently, there
are four hydrogen atoms on the reactant side and 2
hydrogen atoms on the product side. Thus, we need to
add a coefficient of 2 in front of water to make the
hydrogen atoms equal.
CH4(g) + O2(g)  CO2(g) + 2H2O(g)
5-16
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Writing Chemical Equations
CH4(g) + O2(g)  CO2(g) + 2H2O(g)
Finally, we look at the oxygen atoms. Currently, there
are 2 oxygen atoms on the reactant side and 4 oxygen
atoms (combined from carbon dioxide and water) on the
product side. Thus, we add a coefficient of 2 in front of
the oxygen gas.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Now the equation is balanced!
5-17
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A General Approach to Balancing
Equations
1.
2.
3.
4.
Identify the reactants and products and write
their correct formulas.
Write a skeletal equation including physical
states.
Change coefficients one at a time until the
atoms of each element are balanced. (Start
with the elements that occur least often in the
equation.)
Make a final check by counting the atoms of
each element on both sides of the equation.
5-18
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Activity: Balancing Equations

Balance the following equations:
1.
2.
3.
4.
Potassium chlorate  potassium chloride +
oxygen
Aluminum acetate reacts with potassium
sulfate to form potassium acetate and
aluminum sulfate
Hexane (C6H14) reacts with oxygen gas to
form carbon dioxide and water
Zinc reacts with hydrochloric acid to form
zinc chloride and hydrogen gas
5-19
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Activity Solutions: Balancing Equations
Potassium chlorate  potassium chloride + oxygen
First, translate the chemical names into chemical
formulas:
1.
Potassium chlorate  potassium chloride + oxygen
K+
ClO3−
K+
Cl−
O2
(diatomic)
KClO3(s)  KCl(aq)
+ O2(g)
Next, balance the chemical equation:
2KClO3(s)  2KCl(aq) + 3O2(g)
5-20
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Activity Solutions: Balancing Equations
Aluminum acetate reacts with potassium sulfate to form
potassium acetate and aluminum sulfate
Aluminum acetate + potassium sulfate 
Al(C2H3O2)3 +
K2SO4 
potassium acetate + aluminum sulfate
KC2H3O2
+
Al2(SO4)3
2.
Al(C2H3O2)3(aq) + K2SO4(aq) 
KC2H3O2(aq) + Al2(SO4)3(s)
Balanced:
2Al(C2H3O2)3(aq) + 3K2SO4(aq) 
6KC2H3O2(aq) + Al2(SO4)3(s)
5-21
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Activity Solutions: Balancing Equations
3.
Hexane (C6H14) reacts with oxygen gas to form carbon
dioxide and water
Hexane was given. Oxygen gas is diatomic. Carbon
dioxide has 1 carbon atom and 2 oxygen atoms (using
the Greek prefixes) and water has 2 hydrogen atoms
and 1 oxygen atom.
C6H14(l) + O2(g)  CO2(g) + H2O(g)
Balanced:
C6H14(l) + 9.5O2(g)  6CO2(g) + 7H2O(g)
Fractional coefficients are not acceptable, so we multiply
all coefficients by 2:
2C6H14(l) + 19O2(g)  12CO2(g) + 14H2O(g)
5-22
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Activity Solutions: Balancing Equations
4.
Zinc reacts with hydrochloric acid to form
zinc chloride and hydrogen gas
Zinc + Hydrochloric acid  Zinc chloride + hydrogen
Zn(s) + HCl(aq)  ZnCl2(aq) + H2(g)
Balanced:
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
5-23
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Activity: Precipitation Reactions

Jennifer mixes solutions of cadmium(II)
nitrate, Cd(NO3)2, and sodium sulfide,
Na2S. She obtains an orange precipitate
that is the pigment known as cadmium
orange. Identify the precipitate, and
write a balanced chemical equation for
the reaction she carried out.
Cd(NO3)2(aq) + Na2S(aq) 
5-24
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Activity Solution: Precipitation Reactions
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The starting solutions contain Cd2+(aq),
NO3−(aq), Na+(aq), and S2−(aq).
The solubility rules say that most compounds of
sodium are soluble and all nitrates are soluble,
so the precipitate must be formed between
cadmium(II) ions and sulfide ions.
By matching the positive and negative charges
to get electrical neutrality, we can determine
the formula of the precipitate to be CdS(s).
5-25
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Activity Solution: Precipitation Reactions
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The solution contains Na+(aq) and
NO3−(aq) ions in a 1:1 ratio, so we write
NaNO3(aq).
We can write the skeletal equation for
this reaction as:
Cd(NO3)2(aq) + Na2S(aq)  CdS(s) + NaNO3(aq)

Balanced, the equation is:
Cd(NO3)2(aq) + Na2S(aq)  CdS(s) + 2NaNO3(aq)
5-26
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Extra Activity: Precipitation Reactions
Given that a precipitate
forms in this reaction,
predict the products and
their physical states, and
balance the equation.
BaCl2(aq) + Na2CrO4(aq)  ?
BaCl2(aq) + Na2CrO4(aq) 
BaCrO4(s) + 2NaCl(aq)
Figure from p. 205
5-27
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Activity: Acid Base Neutralizations

Calcium oxide is the white powder, lime.
When added to water, it makes slaked
lime, which is a solution of the base
calcium hydroxide. If sulfuric acid is
added to slaked lime, what products
form? Write the balanced chemical
equation for this reaction.
5-28
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Activity Solution: Acid-Base
Neutralizations
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The starting solutions contain H+(aq), SO42−(aq),
Ca2+(aq), and OH−(aq).
The solubility rules say that most sulfate
compounds are soluble, unless sulfate is
combined with calcium. Thus, calcium sulfate is
a precipitate. Hydrogen ions and hydroxide
ions form water, H2O(l).
By matching the positive and negative charges
to get electrical neutrality, we can determine the
formula of the precipitate to be CaSO4(s).
5-29
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Activity Solution: Acid-Base
Neutralizations

The solution contains no spectator ions.
We can write the skeletal equation for
this reaction as:
H2SO4(aq) + Ca(OH)2(aq)  CaSO4(s) + H2O(l)

Balanced, the equation is:
H2SO4(aq) + Ca(OH)2(aq)  CaSO4(s) + 2H2O(l)
5-30
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Activity: Double-Displacement Reactions
Acid-Base Neutralization Reactions

Complete and balance the following equation:
HCl(aq) + Mg(OH)2(s)  ?
2HCl(aq) + Mg(OH)2(s) 
MgCl2(aq) + 2H2O(l)
Figure 11.6
5-31
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Combustion Reactions
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A combustion reaction is a 5th type of reaction that
does not always fall into one of the earlier categories.
In a combustion reaction, a substance reacts with
oxygen in a reaction that burns to produce a flame.
Figure 5.8F
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Figure 5.31
5-32
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Figure 5.30
Figure 5.8F
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
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5-33
Combustion of Hydrocarbons

The most common combustion reactions
involve the burning of a hydrocarbon, or a
hydrocarbon containing oxygen. The
products are always carbon dioxide and
water.
2C8H18(g) + 25O2(g)  16CO2(g) + 18H2O(g)
5-34
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5.5 Representing Reactions in
Aqueous Solution

When reactions occur in water, soluble ionic
compounds exist as separate ions, not grouped
together with each other as might seem so from the
equation, sometimes called a molecular equation:
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
NaCl(aq)
exists in
solution as
separate
ions
Figure 3.5
5-35
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Representing Reactions in Aqueous
Solution
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An ionic equation shows soluble ionic
compounds as separate ions:
Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq) 
AgCl(s) + Na+(aq) + NO3(aq)
Insoluble compounds do not exist as ions.
Spectator ions are ions that occur on both sides of the
equation.
5-36
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Ionic Equations and Spectator Ions

What are the spectator ions in the ionic
equation?
Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq) 
AgCl(s) + Na+(aq) + NO3(aq)

Since the spectator ions do not change, we can leave
them out of the net ionic equation:
Ag+(aq) + Cl(aq)  AgCl(s)
5-37
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Activity: Representing Reactions in
Aqueous Solution

Barium sulfate, used in the white pigment
lithopone, can be prepared by mixing
solutions of barium chloride and sodium
sulfate. Write balanced molecular, ionic, and
net ionic equations for this reaction.
Figure 5.26
5-38
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Activity Solution: Representing
Reactions in Aqueous Solution
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Both reactants are strong electrolytes, so they
dissociate into the ions Ba2+(aq), Cl-(aq),
Na+(aq) and SO42-(aq) in solution.
Possible formulas for the precipitate are
BaCl2, Na2SO4, NaCl, or BaSO4.
The first two choices can be eliminated
because they are starting reactants.
The solubility rules (Table 5.4) indicate that
NaCl is soluble and BaSO4 is insoluble.
Thus, we can write the molecular equation:
BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq)
5-39
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Activity Solution: Representing
Reactions in Aqueous Solution
BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq)
Since BaCl2, Na2SO4, and NaCl are all strong
electrolytes in solution, we can write their
formulas as separate ionic formulas, giving the
following ionic equation:
Ba2+(aq) + 2Cl−(aq) + 2Na+(aq) + SO42−(aq) 
BaSO4(s) + 2Na+(aq) + 2Cl−(aq)
5-40
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Activity Solution: Representing
Reactions in Aqueous Solution
Ba2+(aq) + 2Cl−(aq) + 2Na+(aq) + SO42−(aq) 
BaSO4(s) + 2Na+(aq) + 2Cl−(aq)
Finally, we note that Na+ and Cl− occur on both
sides of the equation, so they are spectator ions.
They can be eliminated, giving us the following
net ionic equation:
Ba2+(aq) + SO42−(aq)  BaSO4(s)
5-41
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Activity: Spectator Ions and Net Ionic
Equations

Identify the spectator ions and write the net ionic
equations from the following molecular equations:
1) Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s)
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)
2) 2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
2Na(s) + 2H2O(l)  2Na+(aq) + 2OH−(aq) + H2(g)
Figure from p. 206
5-42
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Activity: Net Ionic Equations

What is the net ionic equation for the reaction of
KCl(aq) and NaNO3(aq)?
Ionic “equation”
K+(aq) + Cl−(aq) + Na+(aq) + NO3−(aq)  K+(aq) + NO3−(aq) + Na+(aq) + Cl−(aq)
No reaction occurs!
Figure 5.7
5-43
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