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Chapter 5 Chemical Reactions and Quantities 5.3
Types of Reactions
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Type of Reactions
Chemical reactions can be classified as § Combination reactions. § Decomposition reactions.
§ Single Replacement reactions.
§ Double Replacement reactions.
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Combination
In a combination reaction, § Two or more elements (or simple compounds)
combine to form one product
+
2Mg(s) + O2(g) 2MgO(s) 2Na(s) + Cl2(g) 2NaCl(s)
SO3(g) + H2O(l) H2SO4(aq)
A B A B
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Formation of MgO
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Decomposition
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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Decomposition of HgO
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Learning Check
Classify the following reactions as 1) combination or 2) decomposition:
___A. H2(g) + Br2(g) 2HBr(l)
___B. Al2(CO3)3(s) Al2O3(s) + 3CO2(g)
___C. 4Al(s) + 3C(s) Al4C3(s)
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Solution
Classify the following reactions as 1) combination or 2) decomposition:
1 A. H2(g) + Br2(g) 2HBr(l)
2 B. Al2(CO3)3(s) Al2O3(s) + 3CO2(g)
1 C. 4Al(s) + 3C(s) Al4C3(s)
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Single Replacement
In a single replacement reaction, § One element takes the place of a different element in
a reacting compound.
Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)
Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)
2
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Zn and HCl is a Single Replacement Reaction
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Double Replacement
In a double replacement, § Two elements in the reactants exchange places.
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g)
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Example of a Double Replacement
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Learning Check
Classify the following reactions as 1) single replacement 2) double replacement
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)
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Solution
Classify the following reactions as 1) single replacement 2) double replacement
1 A. 2Al(s) + 3H2SO4(aq) Al2(SO4)3(s) + 3H2(g)
2 B. Na2SO4(aq) + 2AgNO3(aq)
Ag2SO4(s) + 2NaNO3(aq)
1 C. 3C(s) + Fe2O3(s) 2Fe(s) + 3CO(g)
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Learning Check
Identify each reaction as 1) combination 2) decomposition 3) single replacement 4) double replacement
A. 3Ba(s) + N2(g) Ba3N2(s) B. 2Ag(s) + H2S(aq) Ag2S(s) + H2(g) C. SiO2(s) + 4HF(aq) SiF4(s) + 2H2O(l) D. PbCl2(aq) + K2SO4(aq) 2KCl(aq) + PbSO4(s) E. K2CO3(s) K2O(aq) + CO2(g)
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Solution
1 A. 3Ba(s) + N2(g) 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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Chapter 5 Chemical Reactions and Quantities
5.4 Oxidation-Reduction Reactions
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Oxidation and Reduction
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)
3
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An oxidation-reduction reaction § Transfers electrons from one reactant to another.
A Loss of Electrons is Oxidation (LEO) Zn(s) Zn2+(aq) + 2e-
A Gain of Electrons is Reduction (GER) Cu2+(aq) + 2e- Cu(s)
Electron Loss and Gain
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Oxidation and Reduction
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Zn and Cu2+
oxidation Zn(s) Zn2+(aq) + 2e- Silvery metal
reduction Cu2+(aq) + 2e- Cu(s)
Blue orange
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Electron Transfer from Zn to Cu2+
Oxidation: electron loss
Reduction: electron gain
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Identify each of the following as 1) oxidation or 2) reduction:
__A. Sn(s) Sn4+(aq) + 4e−
__B. Fe3+(aq) + 1e− Fe2+(aq)
__C. Cl2(g) + 2e− 2Cl-(aq)
Learning Check
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Identify each of the following as 1) oxidation or 2) reduction:
1 A. Sn(s) Sn4+(aq) + 4e−
2 B Fe3+(aq) + 1e− Fe2+(aq)
2 C. Cl2(g) + 2e− 2Cl-(aq)
Solution
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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− oxidation
Fluorine atoms gain electrons to form fluoride ions. F2(s) + 2e- 2F−(s) reduction
Writing Oxidation and Reduction Reactions
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In light-sensitive sunglasses, UV light initiates an oxidation-reduction reaction. uv light
Ag+ + Cl− Ag + Cl A. Which reactant is oxidized? B. Which reactant is reduced?
Learning Check
27
In light-sensitive sunglasses, UV light initiates an oxidation-reduction reaction. uv light
Ag+ + Cl− Ag + Cl A. Which reactant is oxidized?
Cl − Cl− Cl + 1e− B. Which reactant is reduced?
Ag+ Ag+ + 1e− Ag
Solution
4
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Learning Check
Identify the substances that are oxidized and reduced in each of the following reactions: A. Mg(s) + 2H+(aq) Mg2+(aq) + H2(g) B. 2Al(s) + 3Br2(g) 2AlBr3(s)
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Solution
A. Mg is oxidized Mg(s) Mg2+(aq) + 2e−
H+ is reduced 2H+ + 2e− H2 B. Al is oxidized 2Al 2Al3+ + 6e−
Br is reduced 2Br + 2e− 2Br−
TYPES OF CHEMICAL REACTIONS • Chemical reactions are often classified into categories
according to characteristics of the reactions. The following is a useful classification scheme:
31
Chapter 5 Chemical Reactions and Quantities 5.5
The Mole
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Collection Terms
A collection term states a specific number of items. § 1 dozen donuts = 12 donuts
§ 1 ream of paper = 500 sheets
§ 1 case = 24 cans
33
A mole is a collection that contains § The same number of particles as there are carbon
atoms in 12.0 g of carbon 12C. § 6.02 x 1023 atoms of an element (Avogadro’s number). 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
A Mole of Atoms
34
A mole § Of a covalent compound has Avogadro’s number of
molecules. 1 mole CO2 = 6.02 x 1023 CO2 molecules 1 mole H2O = 6.02 x 1023 H2O molecules
§ Of an ionic compound contains Avogadro’s number of formula units. 1 mole NaCl = 6.02 x 1023 NaCl formula units 1 mole K2SO4 = 6.02 x 1023 K2SO4 formula units
A Mole of a Compound
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Particle in One-Mole Samples
TABLE 5.3
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Avogadro’s number 6.02 x 1023 can be written as an equality and two conversion factors.
Equality: 1 mole = 6.02 x 1023 particles
Conversion Factors: 6.02 x 1023 particles and 1 mole 1 mole 6.02 x 1023 particles
Avogadro’s Number
5
37
Using Avogadro’s Number
Avogadro’s number is used to convert moles of a substance to particles. How many Cu atoms are in 0.50 mole Cu? 0.50 mole Cu x 6.02 x 1023 Cu atoms
1 mole Cu = 3.0 x 1023 Cu atoms
38
Using Avogadro’s Number
Avogadro’s number is used to convert particles of a substance to moles. How many moles of CO2 are in 2.50 x 1024 molecules CO2? 2.50 x 1024 molecules CO2 x 1 mole CO2
6.02 x 1023 molecules CO2
= 4.15 moles CO2
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1. The number of atoms in 2.0 moles Al is 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 A. 1.0 mole S atoms B. 3.0 moles S atoms C. 1.1 x 1048 moles S atoms
Learning Check
40
C. 1.2 x 1024 Al atoms 2.0 moles Al x 6.02 x 1023 Al atoms
1 mole Al
B. 3.0 mole S atoms 1.8 x 1024 S atoms x 1 mole S
6.02 x 1023 S atoms
Solution
41
Subscripts and Moles
The subscripts in a formula give § The relationship of atoms in the formula. § The moles of each element in 1 mole of compound.
Glucose C6H12O6
In 1 molecule: 6 atoms C 12 atoms H 6 atoms O In 1 mole: 6 moles C 12 moles H 6 moles O
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Subscripts State Atoms and Moles
1 mole C9H8O4 = 9 moles C 8 moles H 4 moles O
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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 8 moles H 4 moles O 1 mole C9H8O4 1 mole C9H8O4 1 mole C9H8O4 and 1 mole C9H8O4 1 mole C9H8O4 1 mole C9H8O4 9 moles C 8 moles H 4 moles O
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Learning Check
A. How many moles O are in 0.150 mole aspirin C9H8O4? B. How many O atoms are in 0.150 mole aspirin C9H8O4?
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Solution
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? 0.150 mole C9H8O4 x 4 moles O x 6.02 x 1023 O atoms
1 mole C9H8O4 1 mole O subscript Avogadro’s
factor Number
= 3.61 x 1023 O atoms
6
46
Chapter 5 Chemical Reactions and Quantities
5.6 Molar Mass
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The molar mass is § The mass of one mole of a substance. § The atomic mass of an element expressed in grams.
Molar Mass
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Give the molar mass to the nearest 0.1 g. A. K = ________
B. Sn = ________
Learning Check
49
Give the molar mass to the nearest 0.1 g. A. K = 39.1 g
B. Sn = 118.7 g
Solution
50
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 of Moles
Atomic Mass Total Mass
Ca 1 40.1 g/mole 40.1 g
Cl 2 35.5 g/mole 71.0 g
CaCl2 111.1 g
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Molar Mass of K3PO4
Determine the molar mass of K3PO4 to 0.1 g.
Element Number of Moles
Atomic Mass Total Mass in K3PO4
K 3 39.1 g/mole 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
52
One-Mole Quantities
32.1 g 55.9 g 58.5 g 294.2 g 342.3 g
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Calculate the molar mass to the nearest 0.1g A. K2O = ______ g
B. Al(OH)3 = ______ g
Learning Check
54
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 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
Solution
7
55
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
Learning Check
56
Prozac, C17H18F3NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac? 3) 309 g/mole
17C (12.0 g) + 18H (1.0 g) + 3F (19.0 g) + 1N (14.0 g) + 1 O (16.0 g) =
204 g + 18 g + 57.0 g + 14.0 + 16.0 g
Solution
57
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
Conversion Factors from Molar Mass
58
Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid.
Learning Check
59
Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass factors for acetic acid. 1 mole of acetic acid = 60.0 g acetic acid 1 mole acetic acid and 60.0 g acetic acid 60.0 g acetic acid 1 mole acetic acid
Solution
60
§ Mole factors are used to convert between the grams of a substance
and the number of moles.
Calculations Using Molar Mass
Grams
Mole factor Moles
61
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 = 81.0 g Al
1 mole Al mole factor for Al
Calculating Grams from Moles
62
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?
Learning Check
63
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
Solution