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Chapter 6
Chemical Reactions

Classification and Mass Relationship
1
Physical and Chemical Change
In a physical change,
 The identity and composition of the
substance do not change.
 The state can change or the material can
be torn into smaller pieces.
In a chemical change,
 New substances form with different
compositions and properties.
 A chemical reaction takes place.
2
Physical and Chemical Change
3
Some Examples of Chemical and
Physical Changes
4
Learning Check
Classify each of the following as a
1) physical change or 2) chemical change
A. ____ Burning a candle.
B. ____ Ice melting on the street.
C. ____ Toasting a marshmallow.
D. ____ Cutting a pizza.
E. ____ Polishing a silver bowl.
5
Solution
Classify each of the following as a
1) physical change or 2) chemical change
A. 2 Burning a candle.
B. 1 Ice melting on the street.
C. 2 Toasting a marshmallow.
D. 1 Cutting a pizza.
E. 2 Polishing a silver bowl.
6
Chemical Reaction


In a chemical
reaction, a
chemical change
produces one or
more new
substances.
During a reaction,
old bonds are
broken and new
bonds are formed.
7
Chemical Reaction
 In a chemical reaction,
atoms in the reactants
are rearranged to form
one or more different
substances.
 In this reaction, Fe and
O2 react to form rust
(Fe2O3).
4Fe + 3O2
2Fe2O3
8
Writing a Chemical Equation
 A chemical equation
 Shows the chemical formulas of the reactants to the
left of an arrow and the products on the right.
Reactants
Products
MgO +
C
CO + Mg
 Can be read in words. “Magnesium oxide reacts
with carbon to form carbon monoxide and
magnesium.”
9
Symbols Used in Equations

Symbols used
in equations
show the states
of the reactants
and products
and the
reaction
conditions.
10
Quantities in A Chemical Reaction
4 NH3 + 5 O2
4 NO + 6 H2O
Four molecules of NH3 react with five
molecules of O2 to produce four molecules
of NO and six molecules of H2O.
or
Four moles of NH3 react with 5 moles of O2
to produce four moles of NO and six moles
of H2O.
11
Law of Conservation of Mass
In any ordinary chemical reaction, matter is
not created nor destroyed.
+
+
H2
+ Cl2
2 HCl=
Total Mass
2(1.0) + 2(35.5)
2(36.5)
73.0 g
=
73.0 g

12
6.2 Balancing a Chemical
Equation

A chemical equation is balanced when there
are the same numbers of each type of atom
on both sides of the equation.
Al
+
S
2Al + 3S
Al2S3 Not Balanced
Al2S3
Balanced
13
Using Coefficients to Balance
 To balance an equation, place coefficients in front of
the appropriate formulas.
4 NH3 + 5 O2
4 NO + 6 H2O
 Check the balance by counting the atoms of each
element in the reactants and the products.
4 N (4 x 1N)
=
4 N (4 x 1N)
12 H (4 x 3H) =
12 H (6 x 2H)
10 O (5 x 2O) =
10 O (4O + 6O)
14
Learning Check
Check the balance of atoms in the following:
Fe3O4 + 4 H2
3 Fe + 4 H2O
A. Number of H atoms in products.
1) 2
2) 4
3) 8
B. Number of O atoms in reactants.
1) 2
2) 4
3) 8
C. Number of Fe atoms in reactants.
1) 1
2) 3
3) 4
15
Solution
Fe3O4 + 4 H2
A.
3 Fe + 4 H2O
Number of H atoms in products.
3) 8 (4H2O)
B. Number of O atoms in reactants.
2) 4 (Fe3O4)
C. Number of Fe atoms in reactants.
2) 3 (Fe3O4)
16
Balancing with Polyatomic Ions



Polyatomic ions can be balanced as a unit when
they appear on both sides.
Pb(NO3)2 + NaCl
NaNO3 + PbCl2
Balance NO3- as a unit
Pb(NO3)2 + NaCl
2NaNO3 + PbCl2
2 NO3–
=
2 NO3–
Balance Na (or Cl)
Pb(NO3)2 + 2NaCl
2NaNO3 + PbCl2
2Na+ = 2Na+
2Cl–
= 2Cl–
17
Learning Check
Balance each equation. The coefficients in the
answers are read from left to right.
__Mg + __N2
1) 1, 3, 2
B.__Al
__Mg3N2
2) 3, 1, 2
+ __Cl2
1) 3, 3, 2
2) 1, 3, 1
3) 3, 1, 1
__AlCl3
3) 2, 3, 2
18
Solution
A. 3) 3, 1, 1
3 Mg + 1 N2
1 Mg3N2
B. 3) 2, 3, 2
2 Al + 3 Cl2
2 AlCl3
19
Collection Terms




A collection term
indicates a specific
number of items.
For example, 1
dozen doughnuts
contains 12
doughnuts.
1 ream of paper
means 500 sheets.
1 case is 24 cans.
20
A Mole



A mole contains 6.02 x 1023 particles, which is
the number of carbon atoms in 12.01 g of
carbon.
1 mole C = 6.02 x 1023 C atoms
The number 6.02 x 1023 is known as Avogadro’s
number.
One mole of any element contains Avogadro’s
number of atoms.
1 mole Na = 6.02 x 1023 Na atoms
1 mole Au = 6.02 x 1023 Au atoms
21
A Mole of Molecules



Avogadro’s number is also the number of
molecules and formula units in one mole of a
compound.
One mole of a covalent compound contains
Avogadro’s number of molecules.
1 mole CO2 = 6.02 x 1023 CO2 molecules
1 mole H2O = 6.02 x 1023 H2O molecules
One mole of an ionic compound contains
Avogadro’s number of formula units.
1 mole NaCl = 6.02 x 1023 NaCl formula units
22
Samples of One Mole Quantities
23
Learning Check
A. Calculate the number of atoms in 2.0 moles of Al.
1) 2.0 Al atoms
2) 3.0 x 1023 Al atoms
3) 1.2 x 1024 Al atoms
B. Calculate the number of moles of S in 1.8 x 1024 S.
1) 1.0 mole S atoms
2) 3.0 mole S atoms
3) 1.1 x 1048 mole S atoms
24
Solution
A. Calculate the number of atoms in 2.0 moles of Al.
3) 1.2 x 1024 Al atoms
2.0 moles Al x 6.02 x 1023 Al atoms
1 mole Al
B. Calculate the number of moles of S in 1.8 x 1024 S.
2) 3.0 mole S
1.8 x 1024 S atoms x 1 mole S
6.02 x 1023 S atoms
25
Molar Mass


The mass of
one mole is
called molar
mass.
The molar
mass of an
element is the
atomic mass
expressed in
grams.
26
Learning Check
Give the molar mass to the nearest 0.1 g.
A. 1 mole of K atoms
=
________
B. 1 mole of Sn atoms =
________
27
Solution
Give the molar mass to the nearest 0.1 g.
A. 1 mole of K atoms
=
39.1 g
B. 1 mole of Sn atoms
=
118.7 g
28
Molar Mass of CaCl2

For a compound, the molar mass is the sum of the
molar masses of the elements in the formula. We
calculate the molar mass of CaCl2 to the nearest
0.1 g as follows.
Element
Number Atomic Mass Total Mass
of Moles
Ca
Cl2
CaCl2
1
2
40.1 g/mole 40.1 g
35.5 g/mole 71.0 g
111.1 g
29
Molar Mass of K3PO4
Determine the molar mass of K3PO4 to 0.1 g.
Element
Number Atomic Mass Total Mass
of Moles
in K3PO4
K
P
O
K3PO4
3
1
4
39.1 g/mole 117.3 g
31.0 g/mole 31.0 g
16.0 g/mole 64.0 g
212.3 g
30
One-Mole Quantities
32.1 g
55.9 g
58.5 g
294.2 g
342.3 g
31
Learning Check
A. 1 mole of K2O
= ______g
B. 1 mole of antacid Al(OH)3
= ______g
32
Solution
A. 1 mole of K2O
2 moles K (39.1 g/mole) + 1 mole O (16.0 g/mole)
78.2 g
+ 16.0 g
= 94.2 g
B. 1 mole of antacid Al(OH)3
1 mole Al (27.0 g/mole) + 3 moles O (16.0 g/mole)
+ 3 moles H (1.0 g/mole)
27.0 g
+ 48.0 g + 3.0 g
= 78.0 g
33
Learning Check
Prozac, C17H18F3NO, is an antidepressant that
inhibits the uptake of serotonin by the brain.
What is the molar mass of Prozac?
1) 40.0 g/mole
2) 262 g/mole
3) 309 g/mole
34
Solution
Prozac, C17H18F3NO, is a widely used antidepressant
that inhibits the uptake of serotonin by the brain.
What is the molar mass of Prozac?
3) 309 g/mole
17C (12.0) + 18H (1.0) + 3F (19.0) + 1N (14.0) + 1 O (16.0) =
204
+ 18
+
57.0
+
14.0
+ 16.0
35
Molar Mass Factors
Methane CH4 known as natural gas is used in
gas cook tops and gas heaters.
1 mole CH4 =
16.0 g
The molar mass of methane can be written as
conversion factors.
16.0 g CH4
and
1 mole CH4
1 mole CH4
16.0 g CH4
36
Learning Check
Acetic acid C2H4O2 gives the sour taste to vinegar.
Write two molar mass conversion factors for acetic
acid.
37
Solution
Acetic acid C2H4O2 gives the sour taste to vinegar.
Write two molar mass factors for acetic acid.
1 mole of acetic acid
1 mole acetic acid
60.0 g acetic acid
=
60.0 g acetic acid
and
60.0 g acetic acid
1 mole acetic acid
38
Calculations with Molar Mass
 Mole factors are used to convert between the
grams of a substance and the number of moles.
Grams
Mole factor
Moles
39
Calculating Grams from Moles
Aluminum is often used for the structure of
lightweight bicycle frames. How many grams
of Al are in 3.00 moles of Al?
3.00 moles Al x 27.0 g Al
1 mole Al
= 81.0 g Al
mole factor for Al
40
Learning Check
The artificial sweetener aspartame (Nutri-Sweet)
C14H18N2O5 is used to sweeten diet foods, coffee and
soft drinks. How many moles of aspartame are
present in 225 g of aspartame?
41
Solution
Calculate the molar mass of C14H18N2O5.
(14 x 12.0) + (18 x 1.0) + (2 x 14.0) + (5 x 16.0)
= 294 g/mole
Set up the calculation using a mole factor.
225 g aspartame x 1 mole aspartame
294 g aspartame
mole factor(inverted)
= 0.765 mole aspartame
42
Conservation of Mass

In a chemical reaction, the mass of the reactants
is equal to the mass of the products.
2 moles Ag
+ 1 mole S = 1 mole Ag2S
2 (107.9 g)
+ 1(32.0 g) = 1 (247.9 g)
247.9 g reactants
= 247.9 g product
43
Moles in Equations
 We can read the equation in “moles” by
placing the word “moles” between each
coefficient and formula.
4 Fe + 3 O2
2 Fe2O3
4 moles Fe + 3 moles O2
2 moles Fe2O3
223.2g
96 g
319.2
44
Writing Mole-Mole Factors
A mole-mole factor is a ratio of the coefficients
for two substances.
4 Fe + 3 O2
2 Fe2O3
Fe and O2
4 mole Fe and 3 mole O2
3 mole O2
4 mole Fe

Fe and Fe2O3
4 mole Fe and
2 mole Fe2O3
2 mole Fe2O3
4 mole Fe
O2 and Fe2O3
3 mole O2
and 2 mole Fe2O3
2 mole Fe2O3
3 mole O2
45
Learning Check
Consider the following equation:
3 H2 + N2
2 NH3
A. A mole factor for H2 and N2 is
1) 3 mole N2 2) 1 mole N2
1 mole H2
3 mole H2
B. A mole factor for NH3 and H2 is
1) 1 mole H2
2) 2 mole NH3
2 mole NH3
3 mole H2
3) 1 mole N2
2 mole H2
3) 3 mole N2
2 mole NH3
46
Solution
3 H2
+ N2
2 NH3
A. A mole factor for H2 and N2 is
2) 1 mole N2
3 mole H2
B. A mole factor for NH3 and H2 is
2) 2 mole NH3
3 mole H2
47
Calculations with Mole Factors
Consider the following reaction:
4 Fe + 3 O2
2 Fe2O3
How many moles of Fe2O3 are produced when 6.0
moles O2 react?
Use the appropriate mole factor to determine the
moles Fe2O3.
6.0 mole O2 x 2 mole Fe2O3 = 4.0 mole Fe2O3
3 mole O2
48
Mass Calculations
49
Methanol (CH3OH) burns in air according to the equation
2CH3OH + 3O2
2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
molar mass
CH3OH
209 g CH3OH x
moles H2O
grams H2O
molar mass
coefficients
H2O
chemical equation
4 mol H2O
18.0 g H2O
1 mol CH3OH
=
x
x
32.0 g CH3OH
2 mol CH3OH
1 mol H2O
235 g H2O
50
The reaction between H2 and O2
produces 13.1 g of water. How many
grams of O2 reacted?
2H2 + O2
2H2O
?g
13.1 g
51
Calculating the Mass of a Reactant
The reaction between H2 and O2 produces 13.1 g of
water. How many grams of O2 reacted?
2H2
+
O2
2H2O
?g
13.1 g
Plan: g H2O
mole H2O
mole O2
g O2
13.1 g H2O x 1 mole H2O x 1 mole O2 x 32.0 g O2
18.0 g H2O
2 mole H2O
1 mole O2
= 11.6 g O2
52
Percent Yield
You prepared cookie dough to make 5 dozen cookies.
The phone rings and you answer. While you talk, a
sheet of 12 cookies burns. You have to throw them out.
The rest of the cookies are okay. The results of our
baking can be described as follows:
Theoretical yield 60 cookies possible
Actual yield
48 cookies to eat
Percent yield
48 cookies x 100 = 80% yield
60 cookies
53
Percent Yield



The theoretical yield is the maximum amount
of product calculated using the balanced
equation.
The actual yield is the amount of product
obtained when the reaction is run.
Percent yield is the ratio of actual yield
compared to the theoretical yield.
Percent Yield =
Actual Yield (g)
x 100
Theoretical Yield (g)
54
Sample Exercise % Yield
Without proper ventilation and limited
oxygen, the reaction of carbon and oxygen
produces carbon monoxide.
2C + O2
2CO
What is the percent yield if 40.0 g of CO are
produced from the reaction of 30.0 g O2?
55
Sample Exercise % Yield (cont.)
1. Calculate theoretical yield of CO.
30.0 g O2 x 1 mole O2 x 2 mole CO x 28.0 g CO
32.0 g O2 1 mole O2
1 mole CO
= 52.5 g CO (theoretical)
2. Calculate the percent yield.
40.0 g CO (actual)
x 100 = 76.2 % yield
52.5 g CO(theoretical)
56
Learning Check
In the lab, N2 and 5.0 g of H2 are reacted
and produce 16.0 g of NH3. What is the
percent yield for the reaction?
N2(g) + 3H2(g)
2NH3(g)
1) 31.3 %
2) 57%
3) 80.0 %
57
Solution
2) 57%
N2(g) + 3H2(g)
2NH3(g)
5.0 g H2 x 1 mole H2 x 2 moles NH3 x 17.0 g NH3
2.0 g H2
3 moles H2
1 mole NH3
= 28 g NH3 (theoretical)
Percent yield = 16.0 g NH3 x 100 = 57 %
28 g NH3
58
Chapter 6 Chemical Reactions
6.8
59
Types of Reactions





Combination
Decomposition
Single Replacement
Double Replacement
Combustion
60
Combination Reactions


In a combination reaction, two or more
elements or simple compounds combine to
form one product.
A
+ B
AB
Examples
H2
+ Cl2
2HCl
2S
+ 3O2
2SO3
4Fe
+ 3O2
2Fe2O3
61
Combination Reactions

In a combination reaction, magnesium and
oxygen react to form magnesium oxide.
2Mg + O2
2MgO
O2
Mg
MgO
62
Decomposition Reactions

In a decomposition reaction, one substance is
broken down (split) into two or more simpler
substances.
AB
A + B
2HgO
2Hg + O2
2KClO3
2KCl + 3 O2
63
Learning Check
Classify the following reactions as
1) combination or 2) decomposition:
___A. H2 + Br2
2HBr
___B. Al2(CO3)3
Al2O3 + 3CO2
___C. 4 Al + 3C
Al4C3
64
Solution
Classify the following reactions as
1) combination or 2) decomposition:
1 A. H2 + Br2
2HBr
2 B. Al2(CO3)3
Al2O3 + 3CO2
1 C. 4 Al + 3C
Al4C3
65
Single Replacement
 In a single replacement, one element takes the
place of an element in a reacting compound.
A
+ BC
AC
+B
Zn(s) + 2HCl(aq)
ZnCl2(aq) + H2(g)
H2
Zn
HCl
ZnCl2
66
Double Replacement

In a double replacement, the positive ions in
the reacting compounds switch places.
AB
+ CD
AD + CB
AgNO3 + NaCl
AgCl + NaNO3
ZnS
ZnCl2 + H2S
+ 2HCl
67
Example of a Double Replacement

When solutions of sodium sulfate and barium
chloride are mixed, solid BaSO4 is produced.
BaCl2 + Na2SO4
BaSO4 + 2NaCl
BaSO4
68
Learning Check
Classify each of the following reactions as a
1) single replacement or 2) double replacement
__A. 2Al + 3H2SO4
Al2(SO4)3 + 3H2
__B. Na2SO4 + 2AgNO3
Ag2SO4 + 2NaNO3
__C. 3C + Fe2O3
2Fe + 3CO
69
Solution
Classify each of the following reactions as a
1) single replacement or 2) double replacement
1 A. 2Al + 3H2SO4
Al2(SO4)3 + 3H2
2 B. Na2SO4 + 2AgNO3
Ag2SO4 + 2NaNO3
1 C. 3C + Fe2O3
2Fe + 3CO
70
Combustion


In a combustion reaction, a reactant often
containing carbon reacts with oxygen O2.
C + O2
CO2
CH4 + 2O2
CO2 + 2H2O
C3H8 + 5O2
3CO2 + 4H2O
Many combustion reactions utilize fuels that
are burned in oxygen to produce CO2, H2O,
and energy.
71
Learning Check
Balance the combustion equation:
___C5H12 + ___O2
___CO2 + ___H2O
72
Solution
Balance the combustion equation:
1 C5H12 + 8 O2
5 CO2 + 6 H2O
73
Reaction of lead
nitrate with sodium
Iodide
PbI2
74
Class of Chemical Reactions
01

Precipitation Reactions: A process in which an
insoluble solid precipitate drops out of the
solution.

Most precipitation reactions occur when the
anions and cations of two ionic compounds
change partners.
Pb(NO3)2(aq) + 2 NaI(aq)  2 NaNO3(aq) + PbI2(s)
75
Types of Chemical Reactions
02

Acid–Base Neutralization: A process in which an
acid reacts with a base to yield water plus an ionic
compound called a salt.

The driving force of this reaction is the formation
of the stable water molecule.
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
76
Types of Chemical Reactions
03

Oxidation–Reduction (Redox) Reaction: A
process in which one or more electrons are
transferred between reaction partners.

The driving force of this reaction is the decrease in
electrical potential.
4 Fe +
3 O2
2 Fe2O3
77
Reaction of lead
nitrate with sodium
Iodide
PbI2
78
Precipitation Reactions
Precipitate – insoluble solid that separates from solution
precipitate
Pb(NO3)2 (aq) + 2NaI (aq)
PbI2 (s) + 2NaNO3 (aq)
molecular equation
Pb2+ + 2NO3- + 2Na+ + 2I-
PbI2 (s) + 2Na+ + 2NO3-
ionic equation
Pb2+ + 2IPbI2
PbI2 (s)
net ionic equation
Na+ and NO3- are spectator ions
79
Writing Net Ionic Equations
1. Write the balanced equation (molecular equation).
2. Write the ionic equation showing the water soluble compounds
in their ionic forms.
3. Determine precipitate from solubility rules
4. Cancel the spectator ions on both sides of the ionic equation
Write the net ionic equation for the reaction of silver
nitrate with sodium chloride in water.
AgNO3 (aq) + NaCl (aq)
AgCl (s) + NaNO3 (aq)
Ag+ + NO3- + Na+ + Cl-
AgCl (s) + Na+ + NO3-
Ag+ + Cl-
AgCl (s)
80
81
Chapter 6 Chemical Reactions
6.12 Redox Reactions
82
Oxidation and Reduction
Oxidation and reduction
 Are an important type of
reaction.
 Provide us with energy
from food.
 Provide electrical energy
in batteries.
 Occur when iron rusts.
4Fe + 3O2
2Fe2O3
83
Electron Loss and Gain



An oxidation-reduction reaction involves
the transfer of electrons from one reactant
to another.
In oxidation, electrons are lost.
Zn
Zn2+ + 2e- (loss of electrons)
In reduction, electrons are gained.
Cu2+ + 2eCu (gain of electrons)
84
Half-Reactions for OxidationReduction

In the oxidation-reduction reaction of zinc and
copper(II) sulfate, the zinc is oxidized and the
Cu2+ (from Cu2+ SO42-) is reduced.
Zn
Zn2+ + 2eoxidation
Cu2+ + 2eCu
reduction
85
Redox Reactions
02
86
Learning Check
Identify each of the following as an
1) oxidation or a 2) reduction:
__A. Sn
Sn4+ + 4e-
__B. Fe3+ + 1e-
Fe2+
+ 2e-
2Cl-
__C. Cl2
87
Solution
Identify each of the following as an
1) oxidation or a 2) reduction:
1 A.
Sn
Sn4+ + 4e-
2 B.
Fe3+ + 1e-
Fe2+
2 C.
Cl2
+ 2e-
2Cl-
88
Balanced Red-Ox Equations


In a balanced oxidation-reduction equation,
the loss of electrons is equal to the gain of
electrons.
Zn + Cu2+
Zn2+ + Cu
The loss and gain of two electrons is shown
in the separate oxidation and reduction
reactions.
Zn
Zn2+ + 2e- oxidation
Cu2+ + 2eCu
reduction
89
Learning Check
In light-sensitive sunglasses, UV light initiates
an oxidation-reduction reaction.
uv light
Ag+ + ClAg + Cl
A. Which reactant is oxidized?
1) Ag+
2) Cl3) Ag
B. Which reactant is reduced?
1) Ag+
2) Cl3) Cl
90
Solution
In light-sensitive sunglasses, UV light initiates
an oxidation-reduction reaction.
uv light
Ag+ + Cl–
Ag + Cl
A. Which reactant is oxidized
2) Cl–
Cl–
Cl + e–
B. Which reactant is reduced?
1) Ag+
Ag+ + e–
Ag
91
Learning Check
Write the separate oxidation and reduction
half reactions for the following equation.
2Cs + F2
2CsF
92
Solution
Write the separate oxidation and reduction
reactions for the following equation.
2Cs + F2
2CsF
2Cs
F2 + 2e–
2Cs+ + 2e–
oxidation
2Freduction
93
Oxidation with Oxygen


An early definition of oxidation is the
addition of oxygen O2 to a reactant.
A metal or nonmetal is oxidized while the
O2 is reduced to O2-.
4K + O2
2K2O
C + O2
CO2
2SO2 + O2
2SO3
94
Gain and Loss of Hydrogen

In organic and biological reactions,
oxidation involves the loss of hydrogen
atoms and reduction involves a gain of
hydrogen atoms.
oxidation = Loss of H
reduction = Gain of H
CH3OH
H2CO + 2H (loss of H)
Methanol
Formaldehyde
95
Oxidation–Reduction Reactions
02
Oxidation
Is
Loss (of electrons)
Anode Oxidation
Reducing Agent
96
Oxidation–Reduction Reactions
03
Reduction
Is
Gain (of electrons)
Cathode Reduction
Oxidizing Agent
97
Oxidation
Number Rules
Rule Applies to
Statement
1
Elements
The oxidation number of an atom in an
element is zero.
2
Monatomic
ions
The oxidation number of an atom in a
monatomic ion equals the charge of the ion.
3
Oxygen
The oxidation number of oxygen is –2 in
most of its compounds. (An exception is O in
H2O2 and other peroxides, where the
oxidation number is –1.)
98
Oxidation Number Rules
Rule Applies to
Statement
4
Hydrogen
+1, it will be -1 when hydrogen comes with
metal. NaH
5
Halogens
6
Compounds
and ions
Fluorine is –1 in all its compounds. Each of
the other halogens is –1 in binary compounds
unless the other element is oxygen.
The sum of the oxidation numbers of the
atoms in a compound is zero. The sum in a
polyatomic ion equals the charge on the ion.
99
Oxidation numbers of all
the elements in HCO3- ?
HCO3O = -2
H = +1
3x(-2) + 1 + ? = -1
C = +4
100
IF7
Oxidation numbers of all
the elements in the
following ?
F = -1
7x(-1) + ? = 0
I = +7
NaIO3
Na = +1 O = -2
3x(-2) + 1 + ? = 0
I = +5
K2Cr2O7
O = -2
K = +1
7x(-2) + 2x(+1) + 2x(?) = 0
Cr = +6
101
Chapter Summary




Chemical equations must be balanced, that is, the
numbers and kinds of atoms must be the same in
the reactants and the products.
To balance an equation, coefficients are placed
before formulas.
The coefficients in a balanced equation represent
the numbers of moles of reactants and products in a
reaction.
A mole refers to Avogadro’s number (6.022x1023)
of formula units of a substance.
102
Chapter Summary Contd.




One mole of any substance has a mass equal to the
molecular or formula weight of the substance in
grams.
The ratios of coefficients act as mole ratios that
relates amounts of reactants and/or products.
The yield of a reaction is the amount of product
obtained.
The percent yield is the amount of product
obtained divided by the amount theoretically
possible and multiplied by 100.
103
Chapter Summary Contd.



Three common types of reactions of ionic
compounds are:
 Precipitation reactions
 Acid-base neutralization reaction
 Oxidation-reduction reactions
By comparing the oxidation numbers of an atom
before and after reaction, we can tell whether the
atom has gained or lost shares in electrons and thus
whether a redox reaction has occurred.
Oxidation numbers are assigned to atoms in
reactants and products to provide a measure for
whether an atom is neutral, electron-rich, or
electron-poor.
104

End of Chapter Six
105
Personal Response System Questions
for use with
Fundamentals of General, Organic, and Biological
Chemistry, 5th ed.
Media Update Edition
McMurry and Castellion
Chapter 6
106
Assume the mixture of substances in drawing (a)
undergoes a reaction. Which of the drawings (b)-(d)
represents a product mixture consistent with the law
of mass conservation?
1.
2.
3.
(a)

(b)
(c)
4.
All drawings (b)-(d)
Only drawing (b)
Only drawing (c)
Only drawing (d)
(d)
107
Assume the mixture of substances in drawing (a)
undergoes a reaction. Which of the drawings (b)-(d)
represents a product mixture consistent with the law
of mass conservation?
2.
All drawings (b)-(d)
Only drawing (b)
3.
Only drawing (c)
4.
Only drawing (d)
1.
(a)

(b)
(c)
(d)
108
Reaction of A (unshaded spheres) with B (shaded
spheres is shown schematically in the diagram
below. Which equation best describes the reaction?
1.

2.
3.
4.
A + B  AB
4 A + 2 B  4 AB
A2 + B2  A2B
2 A2 + B2  2 A2B
109
Reaction of A (unshaded spheres) with B (shaded
spheres is shown schematically in the diagram
below. Which equation best describes the reaction?
1.

2.
3.
4.
A + B  AB
4 A + 2 B  4 AB
A2 + B2  A2B
2 A2 + B2  2 A2B
110
a CH3CH2OH + b O2  c CO2 + d H2O
When the above equation is balanced, the
coefficients a, b, c, and d are
1.
2.
3.
4.
a=1, b=1, c=1, d=1
a=1, b=2, c=2, d=3
a=1, b=3, c=2, d=3
a=2, b=7, c=4, d=6
111
a CH3CH2OH + b O2  c CO2 + d H2O
When the above equation is balanced, the
coefficients a, b, c, and d are
1.
2.
3.
4.
a=1, b=1, c=1, d=1
a=1, b=2, c=2, d=3
a=1, b=3, c=2, d=3
a=2, b=7, c=4, d=6
112
Which of the following reactions is an acid-base
neutralization reaction?
1.
2.
3.
4.
AgNO3(aq) + NaBr(aq)  AgBr(s) + NaNO3(aq)
2 CH3OH(l) + 3 O2 (g)  2 CO2(g) + 4 H2O(l)
2 Na(s) + Br2(l)  2 NaBr(s)
H2SO4(aq) + 2 KOH(aq)  2 H2O(l) + K2SO4(aq)
113
Which of the following reactions is an acid-base
neutralization reaction?
1.
2.
3.
4.
AgNO3(aq) + NaBr(aq)  AgBr(s) + NaNO3(aq)
2 CH3OH(l) + 3 O2 (g)  2 CO2(g) + 4 H2O(l)
2 Na(s) + Br2(l)  2 NaBr(s)
H2SO4(aq) + 2 KOH(aq)  2 H2O(l) + K2SO4(aq)
114
Predict which of the following solutions will result in
a precipitation reaction.
1.
2.
3.
4.
Cr(NO3)3(aq) + KCl(aq) 
CuCl2(aq) + Na2S(aq) 
NH4Br(aq) + Na2SO4(aq) 
CsOH(aq) + RbCl(aq) 
115
Predict which of the following solutions will result in
a precipitation reaction.
1.
2.
3.
4.
Cr(NO3)3(aq) + KCl(aq) 
CuCl2(aq) + Na2S(aq) 
NH4Br(aq) + Na2SO4(aq) 
CsOH(aq) + RbCl(aq) 
116
Identify the oxidized reactant, the reduced reactant,
the oxidizing agent, and the reducing agent in the
reaction: Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g).
1.
2.
3.
4.
CO(g) is oxidized and is the oxidizing agent and
Fe2O3(s) is reduced and is the reducing agent.
CO(g) is oxidized and is the reducing agent and
Fe2O3(s) is reduced and is the oxidizing agent.
CO(g) is reduced and is the oxidizing agent and
Fe2O3(s) is oxidized and is the reducing agent.
CO(g) is reduced and is the reducing agent and
Fe2O3(s) is oxidized and is the oxidizing agent.
117
Identify the oxidized reactant, the reduced reactant,
the oxidizing agent, and the reducing agent in the
reaction: Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g).
1.
2.
3.
4.
CO(g) is oxidized and is the oxidizing agent and
Fe2O3(s) is reduced and is the reducing agent.
CO(g) is oxidized and is the reducing agent and
Fe2O3(s) is reduced and is the oxidizing agent.
CO(g) is reduced and is the oxidizing agent and
Fe2O3(s) is oxidized and is the reducing agent.
CO(g) is reduced and is the reducing agent and
Fe2O3(s) is oxidized and is the oxidizing agent.
118
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