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6.1 Chemical Equations
► 6.1 Chemical Equations
► 6.2 Balancing Chemical Equations
► 6.3 Avogadro’s Number and the Mole
► 6.4 Gram–Mole Conversions
► 6.5 Mole Relationships and Chemical Equations
► 6.6 Mass Relationships and Chemical Equations
► 6.7 Percent Yield
► 6.8 Classes of Chemical Reactions
► 6.9 Precipitation Reactions and Solubility Guidelines
► 6.10 Acids, Bases, and Neutralization Reactions
► 6.11 Redox Reactions
► 6.12 Recognizing Redox Reactions
► 6.13 Net Ionic Equations
Prentice Hall © 2007
Chapter Six
► Chemical equation: An expression in which
symbols are used to represent a chemical reaction.
► Reactant: A substance that undergoes change in a
chemical reaction and is written on the left side of
the reaction arrow in a chemical equation.
► Product: A substance that is formed in a chemical
reaction and is written on the right side of the
reaction arrow in a chemical equation.
1
► The numbers and kinds of atoms must be the same on
both sides of the reaction arrow.
► Numbers in front of formulas are called coefficients;
they multiply all the atoms in a formula.
► The symbol 2 NaHCO3 indicates two units of sodium
bicarbonate, which contains 2 Na, 2 H, 2 C, and 6 O.
► Substances involved in chemical reactions may be
solids, liquids, gases, or they may be in solution.
► This information is added to an equation by placing the
appropriate symbols after the formulas:
► Solid=(s) Liquid=(l) Gas=(g) Aqueous solution=(aq)
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Chapter Six
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► A polyatomic ion appearing on both sides of an
equation can be treated as a single unit.
Chapter Six
Chapter Six
2
6.2 Balancing Chemical Equations
► Balancing chemical equations can be done using four
basic steps:
► STEP 1: Write an unbalanced equation, using the
correct formulas for all reactants and products.
► STEP 2: Add appropriate coefficients to balance the
numbers of atoms of each element.
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Chapter Six
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► STEP 4: Make sure the coefficients are reduced to
their lowest whole-number values.
► The equation:
2 H2SO4 + 4 NaOH → 2 Na2SO4 + 4 H2O
is balanced, but can be simplified by dividing all
coefficients by 2:
H2SO4 + 2 NaOH → Na2SO4 + 2 H2O
► Hint: If an equation contains a pure element as a
product or reactant it helps to assign that element’s
coefficient last.
► STEP 3: Check the equation to make sure the
numbers and kinds of atoms on both sides of the
equation are the same.
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Prentice Hall © 2007
Chapter Six
6
1
Symbols Used in Equations
Symbols used in
chemical
equations show
-TABLE
Chemical Equations are Balanced
In a balanced
chemical reaction
5.2
• atoms are not
gained or lost.
• the states of the
reactants.
• the number of
reactant atoms is
equal to the
number of product
atoms.
• the states of the
products.
• the reaction
conditions.
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8
A Balanced Chemical Equation
Learning Check
In a balanced chemical equation,
Determine if each equation is balanced or not.
• there must be the same number of each type of atom on the
reactant side and on the product side of a balanced equation.
A. Na(s) + N2(g) Na3N(s)
• numbers called coefficients are used in front of one or more
formulas.
B. C2H4(g) + H2O(l) C2H5OH(l)
Al +
2Al
S
Al2S3
Not Balanced
+ 3S Al2S3
2 Al
3S
Balanced
=
2 Al
=
3S
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10
-Guide
to Balancing a Chemical
Equation
Solution
Determine if each equation is balanced or not.
A. Na(s) + N2(g) Na3N(s)
No. 2 N on reactant side, 1 N on product side.
1 Na on reactant side, 3 Na on product side.
B. C2H4(g) + H2O(l) C2H5OH(l)
Yes. 2 C
= 2C
6H
= 6H
1O
= 1O
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Education, Inc.
-Publishing as Benjamin Cummings
11
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2
Steps in Balancing an Equation
Balancing Chemical Equations
1. Write the equation with the correct formulas.
NH3(g) + O2(g) NO(g) + H2O(g)
To balance the following equation,
Fe3O4(s) + H2(g) Fe(s) + H2O(l)
2. Determine if the equation is balanced.
No, not all atoms are balanced.
• work on one element at a time.
• use only coefficients in front of formulas.
• do not change any subscripts.
Fe:
Fe3O4(s) + H2(g) 3Fe(s) + H2O(l)
O:
Fe3O4(s) + H2(g) 3Fe(s)
+ 4H2O(l)
H:
Fe3O4(s) + 4H2(g) 3Fe(s)
+ 4H2O(l)
3. Balance with coefficients in front of formulas.
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
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4. Check that atoms of each element are equal in reactants and
products.
4 N (4 x 1 N) =
4 N (4 x 1 N)
12 H (4 x 3 H) =
12 H (6 x 2 H)
10 O (5 x 2 O) =
10 O (4 O + 6 O)
Learning Check
Learning Check
Check the balance of atoms in the following.
Balance each equation and list the coefficients in the balanced
equation going from reactants to products:
Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(l)
A. __Mg(s) + __N2(g) A. Number of H atoms in products.
1) 2
2) 4
14
3) 8
1) 1, 3, 2
__Mg3N2(s)
2) 3, 1, 2
3) 3, 1, 1
+ __Cl2(g) __AlCl3(s)
B. Number of O atoms in reactants.
1) 2
2) 4
3) 8
B. __Al(s)
C. Number of Fe atoms in reactants.
1) 1
2) 3
3) 4
1) 3, 3, 22) 1, 3, 1
3) 2, 3, 2
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6.3 Avogadro’s Number and the Mole
Solution
► Molecular weight: The sum of atomic weights of
all atoms in a molecule.
► Formula weight: The sum of atomic weights of all
atoms in one formula unit of any compound.
► Mole: One mole of any substance is the amount
whose mass in grams (molar mass) is numerically
equal to its molecular or formula weight.
► Avogadro’s number: The number of molecules or
formula units in a mole. NA = 6.022 x 1023
A. 3) 3, 1, 1
3Mg(s) + 1N2(g) 1Mg3N2(s)
B. 3) 2, 3, 2
2Al(s)
16
+ 3Cl2(g) 2AlCl3(s)
17
Prentice Hall © 2007
Chapter Six
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3
6.4 Gram – Mole Conversions
► Molar mass
= Mass of 1 mole of a substance.
= Mass of 6.022 x 1023 molecules
of a substance.
= Molecular (formula) weight of
substance in grams.
► Molar mass serves as a conversion factor between
numbers of moles and mass. If you know how many
moles you have, you can calculate their mass; if you
know the mass of a sample, you can calculate the
number of moles.
Prentice Hall © 2007
19
Chapter Six
The molar mass of water is 18.0 g. The conversion
factor between moles of water and mass of water is
18.0 g/mol and the conversion factor between mass
of water and moles of water is 1 mol/18.0 g:
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Collection Terms
Chapter Six
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A Mole of Atoms
A collection term states
a specific number of items.
A mole is a collection that contains
• the same number of particles as there are carbon atoms in
12.0 g of carbon.
• 1 dozen donuts
• 6.02 x 1023 atoms of an element (Avogadro’s number).
= 12 donuts
• 1 ream of paper
= 500 sheets
• 1 case
= 24 cans
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© 2005 by Pearson Education, Inc.
as Benjamin Cummings
1 mole element
Number of Atoms
1 mole C
= 6.02 x 1023 C atoms
1 mole Na
= 6.02 x 1023 Na atoms
1 mole Au
= 6.02 x 1023 Au atoms
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A Mole of a Compound
Avogadro’s Number
Avogadro’s number 6.02 x 1023 can be written as an
equality and two conversion factors.
A mole
• of a covalent compound has Avogadro’s number of
molecules.
1 mole CO2 = 6.02 x 1023 CO2 molecules
Equality:
1 mole = 6.02 x 1023 particles
1 mole H2O = 6.02 x 1023 H2O molecules
Conversion Factors:
6.02 x 1023 particles and
1 mole
• of an ionic compound contains Avogadro’s number of
formula units.
1 mole NaCl
= 6.02 x 1023 NaCl formula units
1 mole
6.02 x 1023 particles
1 mole K2SO4 = 6.02 x 1023 K2SO4 formula units
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4
Using Avogadro’s Number
Using Avogadro’s Number
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.
How many Cu atoms are in 0.50 mole Cu?
How many moles of CO2 are in
2.50 x 1024 molecules CO2?
0.50 mole Cu x 6.02 x 1023 Cu atoms
1 mole Cu
= 3.0 x
1023
2.50 x 1024 molecules CO2 x
1 mole CO2
6.02 x 1023 molecules CO2
Cu atoms
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= 4.15 mole CO2
© 2005 by Pearson Education, Inc.
as Benjamin Cummings
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Learning Check
Solution
1. The number of atoms in 2.0 mole Al is
C. 1.2 x 1024 Al atoms
A. 2.0 Al atoms.
2.0 mole Al x 6.02 x 1023 Al atoms
1 mole Al
B. 3.0 x 1023 Al atoms.
C. 1.2 x 1024 Al atoms.
B. 3.0 mole S atoms
2. The number of moles of S in 1.8 x 1024 atoms S is
1.8 x 1024 S atoms x
A. 1.0 mole S atoms.
B. 3.0 mole S atoms.
1 mole S
6.02 x 1023 S atoms
C. 1.1 x 1048 mole S atoms.
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6.5 Mole Relationships and Chemical
Equations
The coefficients in a balanced chemical equation tell
how many molecules, and thus how many moles, of
each reactant are needed and how many molecules, and
thus moles, of each product are formed. See the
example below:
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Chapter Six
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28
► The coefficients can be put in the form of mole
ratios, which act as conversion factors when setting
up factor-label calculations.
► In the ammonia synthesis, the mole ratio of H2 to N2
is 3:1, the mole ratio of H2 to NH3 is 3:2, and the
mole ratio of N2 to NH3 is 1:2 leading to the
following conversion factors:
(3 mol H2)/(1 mol N2)
(3 mol H2)/(2 mol NH3)
(1 mol N2)/(2 mol NH3)
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Chapter Six
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5
Subscripts and Moles
Factors from Subscripts
The subscripts in a formula show
• the relationship of atoms in the formula.
• the moles of each element in 1 mole of compound.
The subscripts are used to write conversion factors for
moles of each element in 1 mole compound. For aspirin
C9H8O4, the following factors can be written:
Glucose
C6H12O6
In 1 molecule: 6 atoms C 12 atoms H
In 1 mole:
6 mole C 12 mole H
9 mole C
1 mole C9H8O4
and
1 mole C9H8O4
9 mole C
8 mole H
1 mole C9H8O4
4 mole O
1 mole C9H8O4
1 mole C9H8O4
8 mole H
1 mole C9H8O4
4 mole O
6 atoms O
6 mole O
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Learning Check
Solution
A. How many mole 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 mole 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?
0.150 mole C9H8O4 x 4 mole O x 6.02 x 1023 O atoms
1 mole C9H8O4 1 mole O
subscript
factor
33
= 3.61 x 1023 O atoms
6.6 Mass Relationships and
Chemical Equations
Avogadro’s
Number
34
Molar Mass
Mole to mole conversions are carried out using mole
ratios as conversion factors.
The molar mass
• is the mass of one mole of
an element or compound.
• is the atomic mass
expressed in grams.
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Chapter Six
35
© 2005 by Pearson Education, Inc.
as Benjamin Cummings
36
6
Molar Mass from Periodic Table
Learning Check
Molar mass is
the atomic mass
expressed in
grams.
Give the molar mass for each (to the tenths decimal place).
1 mole Ag
1 mole S
- = 107.9 g
g
= 32.07 g
-
A. 1 mole K atoms =
________
B. 1 mol Sn atoms =
________
1 mole
C
= 12.01
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Solution
Molar Mass of K3PO4
Calculate the molar mass of K3PO4.
Give the molar mass for each (to the tenths decimal place).
A.
B.
1 mole K atoms =
Element Number
of Moles
K
3
39.1 g
1 mole Sn atoms = 118.7 g
Atomic Mass
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
K3PO4
212.3 g
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40
Some One-mole Quantities
-One
One--Mole
Learning Check
Quantities
What is the molar mass of each of the following?
A. K2O
B. Al(OH)3
32.1 g
55.9 g
58.5 g
294.2 g
342.2 g
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42
7
Solution
Molar Mass Factors
A. K2O 94.2 g/mole
Molar mass conversion factors
• are written from molar mass.
• relate grams and moles of an element or compound.
2 mole K (39.1 g/mole) + 1 mole O (16.0 g/mole)
78.2 g + 16.0 g = 94.2 g
Example: Write molar mass factors for methane, CH4, used in
gas stoves and gas heaters.
Molar mass:
1 mol CH4 = 16.0 g
B. Al(OH)3 78.0 g/mole
1 mole Al (27.0 g/mole) + 3 mol O (16.0 g/mole)
+ 3 mol H (1.01 g/mole)
27.0 g + 48.0 g + 3.03 g = 78.0 g
Conversion factors:
16.0 g CH4
1 mole CH4
and
1 mole CH4
16.0 g CH4
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Learning Check
Calculations Using Molar Mass
Molar mass factors are used to convert between the grams of a
substance and the number of moles.
-
Grams
-Molar
mass factor
Allyl sulfide C6H10S is a
compound that has the odor of
garlic. How many moles of
C6H10S are in 225 g?
-Moles
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46
Solution
Grams, Moles, and Particles
A molar mass factor and Avogadro’s number convert
• grams to particles.
Calculate the molar mass of C6H10S.
(6 x 12.0) + (10 x 1.01) + (1 x 32.1) = 114.2 g/mole
molar mass
Set up the calculation using a mole factor.
225 g C6H10S x 1 mole C6H10S
114.2 g C6H10S
g
Avogadro’s
number
mole
particles
• particles to grams.
molar mass factor (inverted)
Avogadro’s
number
= 1.97 mole C6H10S
particles
47
molar mass
mole
g
48
8
Learning Check
Solution
How many H2O molecules are in 24.0 g H2O?
How many H2O molecules are in 24.0 g H2O?
1) 4.52 x 1023
3) 8.02 x 1023
2) 1.44 x 1025
24.0 g H2O x 1 mole H2O x 6.02 x 1023 H2O molecules
18.0 g H2O
1 mole H2O
3) 8.03 x 1023
= 8.03 x 1023 H2O molecules
49
► Mole to mass and mass to mole conversions are
carried out using molar mass as a conversion factor.
► Mass to mass conversions are frequently needed,
but cannot be carried out directly.
► Overall, there are four steps for determining mass
relationships among reactants and products.
Prentice Hall © 2007
Mass to mass conversions:
►STEP 1: Write the
balanced chemical equation.
►STEP 2: Choose molar
masses and mole ratios to
convert known information
into needed information.
►STEP 3: Set up the factorlabel expression, and
calculate the answer.
►STEP 4: Estimate or check
the answer using a ballpark
solution.
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Chapter Six
50
Prentice Hall © 2007
Reading Equations In Moles
Consider the following equation:
4 Fe(s)
+ 3 O2(g)
52
Writing Mole-Mole Factors
A mole-mole factor is a ratio of the moles for any two
substances in an equation.
2 Fe2O3(s)
4Fe(s) + 3O2(g)
This equation can be read in “moles” by placing the
word “moles” between each coefficient and formula.
4 moles Fe + 3 moles O2
Chapter Six
2Fe2O3(s)
Fe and O2
4 moles Fe and 3 moles O2
3 moles O2
4 moles Fe
Fe and Fe2O3
4 moles Fe and 2 moles Fe2O3
2 moles Fe2O3 4 moles Fe
O2 and Fe2O3 3 moles O2 and 2 moles Fe2O3
2 moles Fe2O3 3 moles O2
2 moles Fe2O3
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9
Learning Check
Solution
3H2(g) + N2(g)
Consider the following equation:
3 H2(g) + N2(g)
2 NH3(g)
A. A mole-mole factor for H2 and N2 is
1) 3 moles N2
2) 1 mole N2
1 mole H2
3 moles H2
B. A mole-mole factor for NH3 and H2 is
1) 1 mole H2
2) 2 moles NH3
2 moles NH3
3 moles H2
2NH3(g)
A. A mole-mole factor for H2 and N2 is
2) 1 mole N2
3 moles H2
3) 1 mole N2
2 moles H2
B. A mole-mole factor for NH3 and H2 is
2) 2 moles NH3
3 moles H2
3) 3 moles N2
2 moles NH3
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56
-Guide
Calculations with Mole Factors
to Using Mole Factors
How many moles of Fe2O3 can form from 6.0 mole O2?
4Fe(s) + 3O2(g)
Relationship:
2Fe2O3(s)
3 mole O2 = 2 mole Fe2O3
Write a mole-mole factor to determine the moles of Fe2O3.
6.0 mole O2 x 2 mole Fe2O3 = 4.0 mole Fe2O3
3 mole O2
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Learning Check
Solution
How many moles of Fe are needed for the reaction
of 12.0 moles O2?
3) 16.0 moles Fe
4 Fe(s)
12.0 moles O2 x
+
3 O2(g)
2 Fe2O3(s)
1) 3.00 moles Fe
2) 9.00 moles Fe
3) 16.0 moles Fe
59
4 moles Fe = 16.0 moles Fe
3 moles O2
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10
Steps in Finding the Moles and Masses in a
Chemical Reaction
Moles to Grams
Suppose we want to determine the mass (g) of NH3
that can form from 2.50 moles N2.
N2(g) + 3 H2(g)
2 NH3(g)
The plan needed would be
moles N2
moles NH3
grams NH3
The factors needed would be:
mole factor NH3/N2 and the molar mass NH3
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Moles to Grams
Learning Check
The setup for the solution would be:
How many grams of O2 are needed to produce
0.400 mole Fe2O3 in the following reaction?
2.50 mole N2 x 2 moles NH3 x 17.0 g NH3
1 mole N2
1 mole NH3
given
mole-mole factor
4 Fe(s) + 3 O2(g)
molar mass
2 Fe2O3(s)
1) 38.4 g O2
2) 19.2 g O2
= 85.0 g NH3
3) 1.90 g O2
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6.7 Percent Yield
Solution
► The amount of product actually formed in a
chemical reaction is somewhat less than the amount
predicted by theory.
► Unwanted side reactions and loss of product during
handling prevent one from obtaining a perfect
conversion of all the reactants to desired products.
► The amount of product actually obtained in a
chemical reaction is usually expressed as a percent
yield.
2) 19.2 g O2
0.400 mole Fe2O3 x 3 mole O2 x 32.0 g O2= 19.2 g O2
2 mole Fe2O3 1 mole O2
mole factor
molar mass
65
Prentice Hall © 2007
Chapter Six
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11
► Percent yield is defined as:
(Actual yield ÷ Theoretical yield) x 100%
6.8 Classes of Chemical Reactions
► When learning about chemical reactions it is helpful
to group the reactions of ionic compounds into three
general classes: precipitation reactions, acid–base
neutralization reactions, and oxidation–reduction
reactions.
► Precipitation reactions are processes in which an
insoluble solid called a precipitate forms when
reactants are combined in aqueous solution.
► The actual yield is found by weighing the product
obtained.
► The theoretical yield is found by a mass-to-mass
calculation.
Prentice Hall © 2007
Chapter Six
67
Prentice Hall © 2007
Chapter Six
6.9 Precipitation Reactions and
Solubility Guidelines
► Acid–base neutralization reactions are processes
in which H+ ions from an acid react with OH- ions
from a base to yield water. An ionic compound
called a salt is also produced. The “salt” produced
need not be common table salt. Any ionic compound
produced in an acid–base reaction is called a salt.
► Oxidation–reduction reactions, or redox
reactions, are processes in which one or more
electrons are transferred between reaction partners
(atoms, molecules, or ions). As a result of this
transfer, the charges on atoms in the various
reactants change.
► Reaction of aqueous Pb(NO3)2
with aqueous KI gives a
yellow precipitate of PbI2.
► To predict whether a
precipitation reaction will
occur on mixing aqueous
solutions of two ionic
compounds, you must know
the solubility of the potential
products.
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Chapter Six
69
If a potential product does not contain at least one of
the ions listed below, it is probably not soluble and
will precipitate from solution when formed.
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Chapter Six
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6.10 Acids, Bases, and Neutralization
Reactions
► When acids and bases are mixed together in correct
proportion, acidic and basic properties disappear.
► A neutralization reaction produces water and a salt.
HA(aq) + MOH(aq) → H2O(l) + MA(aq)
acid + base
→ water + salt
► The reaction of hydrochloric acid with potassium
hydroxide to produce potassium chloride is an
example:
► HCl(aq) + KOH(aq) → H2O(l) + KCl(aq)
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Chapter Six
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6.11 Redox Reactions
► Oxidation–reduction (redox) reaction: A reaction in
which electrons transfer from one atom to another.
► Oxidation: Loss of one or more electrons by an atom.
► Reduction: Gain of one or more electrons by an atom.
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Chapter Six
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6.12 Recognizing Redox Reactions
Reducing agent:
► Loses one or more electrons
► Causes reduction
► Undergoes oxidation
► Becomes more positive (or less negative)
Oxidizing agent:
► Gains one or more electrons
► Causes oxidation
► Undergoes reduction
► Becomes more negative (or less positive)
Prentice Hall © 2007
► Oxidation and reduction always occur together.
► A substance that is oxidized gives up an electron,
causes reduction, and is called a reducing agent.
► A substance that is reduced gains an electron, causes
oxidation, and is called an oxidizing agent.
► The charge on the reducing agent increases during
the reaction, and the charge on the oxidizing agent
decreases.
► One can determine whether atoms are oxidized or
reduced in a reaction by keeping track of changes
in electron sharing by the atoms. Each atom in a
substance is assigned a value called an oxidation
number or oxidation state.
► The oxidation number indicates whether the atom
is neutral, electron-rich, or electron-poor.
► By comparing the oxidation state of an atom before
and after reaction, we can tell whether the atom has
gained or lost electrons.
75
► Rules for assigning oxidation numbers:
► An atom in its elemental state has an oxidation
number of zero.
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Chapter Six
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► In a molecular compound, an atom usually has the
same oxidation number it would have if it were a
monatomic ion.
► Examples: H often has an oxidation number of +1,
oxygen often has an oxidation number of -2,
halogens often have an oxidation number of -1.
► A monatomic ion has an oxidation number equal to
its charge.
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Chapter Six
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13
6.13 Net Ionic Equations
► For compounds with more than one nonmetal
element, such as SO2, NO, and CO2, the more
electronegative element—oxygen in these
examples—has its preferred negative oxidation
number.
► The less electronegative element is assigned a
positive oxidation number so that the sum of the
oxidation numbers in a neutral compound is 0.
Prentice Hall © 2007
Chapter Six
► Ionic equation: An equation in which ions are
explicitly shown.
► Spectator ion: An ion that appears unchanged on
both sides of a reaction arrow.
► Net ionic equation: An equation that does not
include spectator ions.
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► The coefficients in a balanced chemical equation
represent the numbers of moles of reactants and
products in a reaction.
► Mole ratios relate amounts of reactants and/or
products. Using molar masses and mole ratios in
factor-label calculations relates unknown masses to
known masses or molar amounts.
► The yield 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%.
► Chemical equations must be balanced; the numbers
and kinds of atoms must be the same in both the
reactants and the products.
► To balance an equation, coefficients are placed
before formulas but the formulas themselves cannot
be changed.
► A mole refers to Avogadro’s number of formula
units of a substance. One mole of any substance has
a mass equal to its formula weight in grams.
► Molar masses act as conversion factors between
numbers of molecules and masses in grams.
Chapter Six
Chapter Six
Chapter Summary Cont.
Chapter Summary
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Prentice Hall © 2007
Chapter Six
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Chapter Summary Cont.
► Precipitation reactions are processes in which an
insoluble solid called a precipitate is formed.
► In acid–base neutralization reactions an acid reacts
with a base to yield water plus a salt.
► Oxidation–reduction (redox) reactions are processes
in which one or more electrons are transferred
between reaction partners.
► Oxidation is the loss of electrons by an atom, and
reduction is the gain of electrons by an atom.
► Oxidation numbers are assigned to provide a
measure of whether an atom is neutral, electron-rich,
or electron-poor.
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