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Chapter 5 Chemical Reactions
and Quantities
5.3
Types of Reactions
Type of Reactions
Combination
Chemical reactions can be classified as
§  Combination reactions.
In a combination reaction,
§  Two or more elements (or simple compounds)
combine to form one product
§  Decomposition reactions.
§  Single Replacement reactions.
A
+
B
A
§  Double Replacement reactions.
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Formation of MgO
B
2Mg(s) + O2(g)
2MgO(s)
2Na(s) + Cl2(g)
2NaCl(s)
SO3(g) + H2O(l)
H2SO4(aq)
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Decomposition
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Decomposition of HgO
In a decomposition reaction,
§  One substance splits into two or more simpler
substances.
2HgO(s)
2Hg(l) + O2(g)
2KClO3(s)
2KCl(s) + 3O2(g)
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Learning Check
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Solution
Single Replacement
Classify the following reactions as
1) combination or 2) decomposition:
Classify the following reactions as
1) combination or 2) decomposition:
___A. H2(g) + Br2(g)
2HBr(l)
1 A. H2(g) + Br2(g)
2HBr(l)
___B. Al2(CO3)3(s)
Al2O3(s) + 3CO2(g)
2 B. Al2(CO3)3(s)
Al2O3(s) + 3CO2(g)
___C. 4Al(s) + 3C(s)
Al4C3(s)
1 C. 4Al(s) + 3C(s)
Al4C3(s)
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In a single replacement reaction,
§  One element takes the place of a different element in
a reacting compound.
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Zn(s) + 2HCl(aq)
ZnCl2(aq) + H2(g)
Fe(s) + CuSO4(aq)
FeSO4(aq) + Cu(s)
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Zn and HCl is a Single Replacement
Reaction
Example of a Double Replacement
Double Replacement
In a double replacement,
§  Two elements in the reactants exchange places.
AgNO3(aq) + NaCl(aq)
AgCl(s) + NaNO3(aq)
ZnS(s)
ZnCl2(aq) + H2S(g)
+ 2HCl(aq)
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Learning Check
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Solution
Learning Check
Classify the following reactions as
1) single replacement 2) double replacement
Classify the following reactions as
1) single replacement 2) double replacement
1 A. 2Al(s) + 3H2SO4(aq)
A. 2Al(s) + 3H2SO4(aq)
Al2(SO4)3(s) + 3H2(g)
B. Na2SO4(aq) + 2AgNO3(aq)
Ag2SO4(s) + 2NaNO3(aq)
C. 3C(s) + Fe2O3(s)
2Fe(s) + 3CO(g)
Al2(SO4)3(s) + 3H2(g)
Ag2SO4(s) + 2NaNO3(aq)
2Fe(s) + 3CO(g)
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Ba3N2(s)
3 B. 2Ag(s) + H2S(aq)
Ag2S(s) + H2(g)
4 C. SiO2(s) + 4HF(aq)
SiF4(s) + 2H2O(l)
4 D. PbCl2(aq) + K2SO4(aq)
2KCl(aq) + PbSO4(s)
2 E. K2CO3(s)
K2O(aq) + CO2(g)
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Ba3N2(s)
Ag2S(s) + H2(g)
SiF4(s) + 2H2O(l)
2KCl(aq) + PbSO4(s)
K2O(aq) + CO2(g)
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Oxidation and Reduction
5.4
Oxidation-Reduction Reactions
1 A. 3Ba(s) + N2(g)
3Ba(s) + N2(g)
2Ag(s) + H2S(aq)
SiO2(s) + 4HF(aq)
PbCl2(aq) + K2SO4(aq)
K2CO3(s)
2) decomposition
4) double replacement
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Chapter 5 Chemical Reactions
and Quantities
Solution
Identify each reaction as
1) combination
3) single replacement
A.
B.
C.
D.
E.
2 B. Na2SO4(aq) + 2AgNO3(aq)
1 C. 3C(s) + Fe2O3(s)
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An oxidation-reduction reaction
§  Provides us with energy from food.
§  Provides electrical energy in batteries.
§  Occurs when iron rusts.
4Fe(s) + 3O2(g)
2Fe2O3(s)
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2
Electron Loss and Gain
Zn and Cu2+
Oxidation and Reduction
An oxidation-reduction reaction
§  Transfers electrons from one reactant to another.
A Loss of Electrons is Oxidation
Zn(s)
Zn2+(aq) + 2e-
(LEO)
A Gain of Electrons is Reduction
Cu2+(aq) + 2eCu(s)
(GER)
oxidation
Zn(s)
Silvery metal
Zn2+(aq) + 2ereduction
Cu2+(aq) + 2eBlue
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Electron Transfer from Zn to Cu2+
Learning Check
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Solution
Identify each of the following as
1) oxidation or 2) reduction:
Oxidation: electron loss
Cu(s)
orange
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Identify each of the following as
1) oxidation or 2) reduction:
__A.
Sn(s)
Sn4+(aq) + 4e−
1 A. Sn(s)
Sn4+(aq) + 4e−
__B.
Fe3+(aq) + 1e−
Fe2+(aq)
2 B Fe3+(aq) + 1e−
Fe2+(aq)
__C.
Cl2(g) + 2e−
2Cl-(aq)
2 C. Cl2(g) + 2e−
2Cl-(aq)
Reduction: electron gain
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Writing Oxidation and Reduction
Reactions
Write the separate oxidation and reduction reactions
for the following equation.
2Cs(s) + F2(g)
2CsF(s)
Each cesium atom loses an electron to form cesium
ion.
2Cs(s)
2Cs+(s) + 2e−
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Learning Check
Solution
In light-sensitive sunglasses, UV light initiates
an oxidation-reduction reaction.
uv light
Ag+ + Cl−
Ag + Cl
In light-sensitive sunglasses, UV light initiates
an oxidation-reduction reaction.
uv light
Ag+ + Cl−
Ag + Cl
A. Which reactant is oxidized?
A. Which reactant is oxidized?
Cl −
Cl−
Cl + 1e−
B. Which reactant is reduced?
B. Which reactant is reduced?
Ag+
Ag+ + 1e−
Ag
oxidation
Fluorine atoms gain electrons to form fluoride ions.
F2(s) + 2e2F−(s)
reduction
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Learning Check
Solution
Identify the substances that are oxidized and reduced in
each of the following reactions:
A. Mg is oxidized
H+ is reduced
Mg(s)
2H+ + 2e−
Mg2+(aq) + 2e−
H2
A. Mg(s) + 2H+(aq)
Mg2+(aq) + H2(g)
B. Al is oxidized
Br is reduced
2Al
2Br + 2e−
2Al3+ + 6e−
2Br−
B. 2Al(s) + 3Br2(g)
2AlBr3(s)
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Chapter 5 Chemical Reactions
and Quantities
5.5
The Mole
TYPES OF CHEMICAL REACTIONS
•  Chemical reactions are often classified into categories
according to characteristics of the reactions. The
following is a useful classification scheme:
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Collection Terms
A Mole of Atoms
A collection term states a specific number of items.
A mole is a collection that contains
§  1 dozen donuts
= 12 donuts
§  The same number of particles as there are carbon
§  1 ream of paper = 500 sheets
atoms in 12.0 g of carbon 12C.
§  6.02 x 1023 atoms of an element (Avogadro s number).
§  1 case = 24 cans
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A Mole of a Compound
1 mole CO2 = 6.02 x
= 6.02 x 1023 Au atoms
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equality and two conversion factors.
Equality:
1 mole
CO2 molecules
1 mole H2O = 6.02 x 1023 H2O molecules
§  Of an ionic compound contains Avogadro s number of
= 6.02 x 1023 particles
Conversion Factors:
formula units.
1 mole NaCl
= 6.02 x 1023 Na atoms
1 mole Au
Avogadro s Number
molecules.
1023
= 6.02 x 1023 C atoms
1 mole Na
Avogadro s number 6.02 x 1023 can be written as an
TABLE 5.3
§  Of a covalent compound has Avogadro s number of
Number of Atoms
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Particle in One-Mole Samples
A mole
1 mole element
1 mole C
= 6.02 x 1023 NaCl formula units
6.02 x 1023 particles
1 mole
1 mole K2SO4 = 6.02 x 1023 K2SO4 formula units
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and
1 mole
6.02 x 1023 particles
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Using Avogadro s Number
Using Avogadro s Number
Learning Check
Avogadro s number is used to convert
moles of a substance to particles.
Avogadro s number is used to convert
particles of a substance to moles.
1. The number of atoms in 2.0 moles Al is
How many Cu atoms are in
How many moles of CO2 are in
2.50 x 1024 molecules CO2?
0.50 mole Cu?
2.50 x 1024 molecules CO2 x
0.50 mole Cu x 6.02 x 1023 Cu atoms
1 mole Cu
= 3.0 x 1023 Cu atoms
A. 2.0 Al atoms
B. 3.0 x 1023 Al atoms
C. 1.2 x 1024 Al atoms
2. The number of moles of S in 1.8 x 1024 atoms S is
1 mole CO2
A. 1.0 mole S atoms
6.02 x 1023 molecules CO2
B. 3.0 moles S atoms
= 4.15 moles CO2
C. 1.1 x 1048 moles S atoms
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Solution
Subscripts and Moles
C. 1.2 x 1024 Al atoms
The subscripts in a formula give
§  The relationship of atoms in the formula.
§  The moles of each element in 1 mole of compound.
2.0 moles Al x 6.02 x 1023 Al atoms
1 mole Al
1.8 x
S atoms x
Subscripts State Atoms and Moles
Glucose
C6H12O6
B. 3.0 mole S atoms
1024
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1 mole S
6.02 x 1023 S atoms
In 1 molecule: 6 atoms C 12 atoms H 6 atoms O
In 1 mole:
6 moles C 12 moles H 6 moles O
1 mole C9H8O4
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Factors from Subscripts
Subscripts used for conversion factors
§  Relate moles of each element in 1 mole compound.
§  For aspirin C9H8O4 can be written as:
9 moles C
1 mole C9H8O4
8 moles H
1 mole C9H8O4
4 moles O
1 mole C9H8O4
1 mole C9H8O4
8 moles H
1 mole C9H8O4
4 moles O
9 moles C 8 moles H
4 moles O
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Learning Check
Solution
A. How many moles O are in 0.150 mole aspirin C9H8O4?
A. How many mole O are in 0.150 mole aspirin C9H8O4?
0.150 mole C9H8O4 x 4 moles O
= 0.600 mole O
1 mole C9H8O4
subscript factor
B. How many O atoms are in 0.150 mole aspirin C9H8O4?
B. How many O atoms are in 0.150 mole aspirin C9H8O4?
and
1 mole C9H8O4
9 moles C
=
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0.150 mole C9H8O4 x 4 moles O x 6.02 x 1023 O atoms
1 mole C9H8O4
1 mole O
subscript
factor
Avogadro s
Number
= 3.61 x 1023 O atoms
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Chapter 5 Chemical Reactions and
Quantities
Molar Mass
Learning Check
The molar mass is
§  The mass of one mole of a substance.
§  The atomic mass of an element expressed in grams.
5.6
Molar Mass
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Solution
Give the molar mass to the nearest 0.1 g.
A. K
=
39.1 g
B. Sn
=
118.7 g
Element
Number of
Moles
Atomic Mass
Total Mass
Ca
1
40.1 g/mole
40.1 g
Cl
2
35.5 g/mole
71.0 g
Learning Check
A. K2O =
Atomic Mass
3
39.1 g/mole
Total Mass in
K3PO4
117.3 g
P
1
31.0 g/mole
31.0 g
O
4
16.0 g/mole
64.0 g
212.3 g
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Calculate the molar mass to the nearest 0.1g
A. K2O
2 moles K (39.1 g/mole) + 1 mole O (16.0 g/mole)
78.2 g
+ 16.0 g
= 94.2 g
______ g
B. 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
342.3 g
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Number of
Moles
K
Solution
B. Al(OH)3 = ______ g
294.2 g
Element
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Calculate the molar mass to the nearest 0.1g
58.5 g
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K3PO4
111.1 g
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55.9 g
________
Determine the molar mass of K3PO4 to 0.1 g.
masses of the elements in the formula. We calculate the molar
mass of CaCl2 to the nearest 0.1 g as follows.
CaCl2
32.1 g
________
=
Molar Mass of K3PO4
§  For a compound, the molar mass is the sum of the molar
One-Mole Quantities
=
B. Sn
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Molar Mass of CaCl2
Give the molar mass to the nearest 0.1 g.
A. K
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6
Learning Check
Conversion Factors from Molar
Mass
Solution
Prozac, C17H18F3NO, is an antidepressant that inhibits the uptake of
serotonin by the brain. What is the molar mass of Prozac?
Prozac, C17H18F3NO, is an antidepressant that inhibits the uptake
of serotonin by the brain. What is the molar mass of Prozac?
Methane CH4 known as natural gas is used in gas cook
tops and gas heaters.
1) 40.0 g/mole
3) 309 g/mole
1 mole CH4 =
2) 262 g/mole
17C (12.0 g) + 18H (1.0 g) + 3F (19.0 g)
3) 309 g/mole
204 g + 18 g + 57.0 g + 14.0 + 16.0 g
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Learning Check
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Solution
Acetic acid C2H4O2 gives the sour taste to vinegar. Write two
molar mass conversion factors for acetic acid.
1 mole of acetic acid
1 mole acetic acid
60.0 g acetic acid
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Calculations Using Molar Mass
Acetic acid C2H4O2 gives the sour taste to
vinegar. Write two molar mass factors for
acetic acid.
=
60.0 g acetic acid
and
60.0 g acetic acid
1 mole acetic acid
§  Mole factors are used to convert between the grams of a substance
and the number of moles.
Grams
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Calculating Grams from Moles
3.00 moles Al x
27.0 g Al
1 mole Al
Mole factor
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Learning Check
Aluminum is often used for the structure of
lightweight bicycle frames. How many grams
of Al are in 3.00 moles of Al?
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
+ 1N (14.0 g) + 1 O (16.0 g) =
Moles
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Solution
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?
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
= 81.0 g Al
Set up the calculation using a mole factor.
mole factor for Al
225 g aspartame x 1 mole aspartame
294 g aspartame
mole factor(inverted)
= 0.765 mole aspartame
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