1. Chapter 5 Chemical Reactions
Chemical Changes
Chemical Equations
Balancing a Chemical Equation

2. Physical Properties
The physical properties of a substance are the characteristics we can observe or measure without changing the substance.

3. 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.

4. Physical and Chemical Change

5. Some Examples of Chemical and Physical Changes

6. 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.

7. Solution 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.

8. 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.

9. 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 + 3O Fe2O3

10. Writing 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.”

11. Symbols Used in Equations
Symbols used in equations show the states of the reactants and products and the reaction conditions.

12. Quantities in A Chemical Reaction
4 NH O NO 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.

13. Law of Conservation of Mass
In any ordinary chemical reaction, matter is not created nor destroyed.
H Cl HCl
Total atoms = Total atoms
2 H, 2 Cl 2H, 2 Cl
Total Mass = Total Mass
2(1.0) (35.5) 2(36.5)
73.0 g = g

14. 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 Al2S Not Balanced
2Al S Al2S3 Balanced

15. Using Coefficients to Balance
To balance an equation, place coefficients in front of the appropriate formulas.
4 NH O NO H2O
Check the balance by counting the atoms of each element in the reactants and the products.
4 N (4 x 1N) = N (4 x 1N)
12 H (4 x 3H) = 12 H (6 x 2H)
10 O (5 x 2O) = 10 O (4O + 6O)

16. Steps in Balancing an Equation
Balance one element at a time.
Use only coefficients to balance.
Fe3O H Fe H2O
Fe: Fe3O H Fe H2O
O: Fe3O H Fe H2O
H: Fe3O H Fe H2O

17. Learning Check
Check the balance of atoms in the following: Fe3O H Fe H2O
A. Number of H atoms in products.
1) 2 2) ) 8
B. Number of O atoms in reactants.
C. Number of Fe atoms in reactants.
1) 1 2) ) 4

18. Solution Fe3O4 + 4 H2 3 Fe + 4 H2O A. Number of H atoms in products.
B. Number of O atoms in reactants.
2) 4 (Fe3O4)
C. Number of Fe atoms in reactants.
2) 3 (Fe3O4)

19. 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 NaNO PbCl2
2 NO = NO3-
Balance Na (or Cl)
Pb(NO3)2 + 2NaCl NaNO3 + PbCl2
2Na = Na+
2Cl = Cl-

20. Learning Check Balance each equation. The coefficients in the
answers are read from left to right.
A. __Mg + __N __Mg3N2
1) 1, 3, ) 3, 1, ) 3, 1, 1
B. __Al __Cl __AlCl3
1) 3, 3, 2 2) 1, 3, ) 2, 3, 2

21. Solution
A. 3) 3, 1, 1
3 Mg N Mg3N2
B. 3) 2, 3, 2
2 Al Cl AlCl3

22. Learning Check A. __Fe2O3 + __C __Fe + __CO2
1) 2, 3, 2, ) 2, 3, 4, ) 1, 1, 2, 3
B. __Al + __FeO __Fe __Al2O3
1) 2, 3, 3, ) 2, 1, 1, ) 3, 3, 3, 1
C. __Al + __H2SO4 __Al2(SO4)3 + __H2
1) 3, 2, 1, ) 2, 3, 1, ) 2, 3, 2, 3

23. Solution A. 2) 2, 3, 4, 3 2 Fe2O3 + 3 C 4 Fe + 3 CO2 B. 1) 2, 3, 3, 1
2 Al FeO Fe Al2O3
C. 2) 2, 3, 1, 3
2 Al H2SO Al2(SO4) H2

24. Chapter 5 Chemical Reactions
Types of Reactions

25. Types of Reactions
Chemical reactions are classified into general types:
Combination
Decomposition
Single Replacement
Double Replacement
Combustion

26. Combination Reactions
In a combination reaction, two or more elements or simple compounds combine to form one product.
A B AB
Examples
H Cl2 2HCl
2S O2 2SO3
4Fe O2 2Fe2O3

27. Combination Reactions
In a combination reaction, magnesium and oxygen react to form magnesium oxide.
2Mg + O MgO O2
MgO Mg

28. 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

29. Learning Check Classify the following reactions as
1) combination or 2) decomposition:
___A. H2 + Br HBr
___B. Al2(CO3)3 Al2O3 + 3CO2
___C. 4 Al + 3C Al4C3

30. Solution Classify the following reactions as
1) combination or 2) decomposition:
1 A. H2 + Br HBr
2 B. Al2(CO3)3 Al2O3 + 3CO2
1 C. 4 Al + 3C Al4C3

31. Single Replacement
In a single replacement, one element takes the place of an element in a reacting compound.
A BC AC B
Zn(s) + HCl(aq) ZnCl2(aq) + H2(g) H2
HCl Zn
ZnCl2

32. Double Replacement
In a double replacement, the positive ions in the reacting compounds switch places.
AB CD AD CB
AgNO3 + NaCl AgCl + NaNO3
ZnS HCl ZnCl2 + H2S

33. Example of a Double Replacement
When solutions of sodium sulfate and barium chloride are mixed, solid BaSO4 is produced.
BaCl Na2SO BaSO NaCl
BaSO4

34. Learning Check Classify each of the following reactions as a
1) single replacement or 2) double replacement
__A. 2Al H2SO Al2(SO4) H2
__B. Na2SO4 + 2AgNO Ag2SO4 + 2NaNO3
__C. 3C + Fe2O Fe CO

35. Solution Classify each of the following reactions as a
1) single replacement or 2) double replacement
1 A. 2Al + 3H2SO Al2(SO4)3 + 3H2
2 B. Na2SO4 + 2AgNO Ag2SO4 + 2NaNO3
1 C. 3C + Fe2O Fe + 3CO

36. Combustion
In a combustion reaction, a reactant often containing carbon reacts with oxygen O2.
C + O CO2
CH4 + 2O CO H2O
C3H O CO2 + 4H2O
Many combustion reactions utilize fuels that are burned in oxygen to produce CO2, H2O, and energy.

37. Learning Check Balance the combustion equation:
___C5H ___O ___CO2 + ___H2O

38. Solution
Balance the combustion equation:
1 C5H O CO H2O

39. Summary of Reaction Types

40. Learning Check Identify each reaction as 1) combination
2) decomposition ) combustion
4) single replacement 5) double replacement
A. 3Ba + N Ba3N2
B. 2Ag + H2S Ag2S + H2
C. SiO HF SiF H2O
D. PbCl2 + K2SO KCl + PbSO4
E. K2CO K2O + CO2
F. C2H O CO2 + 2H2O

41. Solution Identify each reaction as 1) combination
2) decomposition ) combustion
4) single replacement 5) double replacement
1 A. 3Ba + N Ba3N2
4 B. 2Ag + H2S Ag2S + H2
5 C. SiO HF SiF H2O
5 D. PbCl2 + K2SO KCl + PbSO4
2 E. K2CO K2O + CO2
3 F. C2H O CO H2O

42. Chapter 5 Chemical Reactions
Oxidation-Reduction Reactions

43. 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 + 3O Fe2O3

44. 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.
Cu e Cu (gain of electrons)

45. Half-Reactions for Oxidation-Reduction
In the oxidation-reduction reaction of zinc and copper(II) sulfate, the zinc is oxidized and the Cu2+ (from Cu2+ SO42-) is reduced.
Zn Zn e- oxidation
Cu e Cu reduction

46. Learning Check Identify each of the following as an
1) oxidation or a 2) reduction:
__A. Sn Sn e-
__B. Fe e- Fe2+
__C. Cl e- 2Cl-

47. Solution Identify each of the following as an
1) oxidation or a 2) reduction:
1 A. Sn Sn e-
2 B. Fe e- Fe2+
2 C. Cl e- 2Cl-

48. Balanced Red-Ox Equations
In a balanced oxidation-reduction equation, the loss of electrons is equal to the gain of electrons.
Zn + Cu Zn2+ + Cu
The loss and gain of two electrons is shown in the separate oxidation and reduction reactions.
Zn Zn2+ + 2e- oxidation
Cu e Cu reduction

49. Learning Check In light-sensitive sunglasses, UV light initiates
an oxidation-reduction reaction.
uv light
Ag+ + Cl Ag Cl
A. Which reactant is oxidized?
1) Ag ) Cl ) Ag
B. Which reactant is reduced?
1) Ag ) Cl ) Cl

50. 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

51. Learning Check
Write the separate oxidation and reduction reactions for the following equation.
2Cs F CsF

52. Solution
Write the separate oxidation and reduction reactions for the following equation.
2Cs F CsF
Cs Cs e- oxidation
F + 1e- F- reduction

53. 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 + O K2O
C + O CO2
2SO2 + O SO3

54. 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 H (loss of H)
Methanol Formaldehyde

55. Summary

56. Learning Check Identify the substances that are oxidized and
reduced in the following reactions.
A. 4Fe + 3O Fe2O3
B. 6Na + N Na3N
C. 2K + I KI

57. Solution 3+ 2- A. 4Fe + 3O2 2Fe2O3 Fe oxidized; O2 reduced 1+ 3-
3+ 2-
A. 4Fe + 3O Fe2O3
Fe oxidized; O2 reduced
1+ 3-
B. 6Na + N Na3N
Na oxidized: N2 reduced
1+ 1-
C. 2K + I KI
K oxidized: I2 reduced

58. The Mole

59. 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.

60. A Mole
A mole contains 6.02 x 1023 particles, which is the number of carbon atoms in g of carbon.
1 mole C = 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 = x 1023 Na atoms
1 mole Au = x 1023 Au atoms

61. 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 = x 1023 CO2 molecules
1 mole H2O = 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

62. Samples of One Mole Quantities

63. Avogadro’s Number Avogadro’s number is written as conversion factors.
6.02 x 1023 particles and 1 mole
1 mole x 1023 particles
The number of molecules in 0.50 mole of CO2 molecules is calculated as
0.50 mole CO2 molecules x 6.02 x 1023 CO2 molecules
1 mole CO2 molecules
= 3.0 x 1023 CO2 molecules

64. Learning Check A. Calculate the number of atoms in 2.0 moles of Al.
1) Al atoms
2) x 1023 Al atoms
3) x 1024 Al atoms
B. Calculate the number of moles of S in 1.8 x 1024 S.
1) mole S atoms
2) mole S atoms
3) x 1048 mole S atoms

65. Solution
A. Calculate the number of atoms in 2.0 moles of Al. 3) x 1024 Al atoms
2.0 moles Al x 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 atoms
1.8 x 1024 S atoms x mole S
6.02 x 1023 S atoms

66. Molar Mass The mass of one mole is called molar mass.
The molar mass of an element is the atomic mass expressed in grams.

67. Learning Check Give the molar mass to the nearest 0.1 g.
A. 1 mole of K atoms = ________
B. 1 mole of Sn atoms = ________

68. 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 = g

69. 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
Cl2 2
35.5 g/mole
71.0 g
CaCl2
111.1 g

70. Molar Mass of K3PO4 Determine the molar mass of K3PO4 to 0.1 g. K 3
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

71. One-Mole Quantities
32.1 g g g g g

72. Learning Check A. 1 mole of K2O = ______g
B. 1 mole of antacid Al(OH)3 = ______g

73. Solution
A. 1 mole of K2O
2 moles K (39.1 g/mole) + 1 mole O (16.0 g/mole)
78.2 g g = 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 g g = g

74. Learning Check
Prozac, C17H18F3NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?
1) g/mole
2) 262 g/mole
3) 309 g/mole

75. 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) =

76. Molar Mass Factors
Methane CH4 known as natural gas is used in gas cook tops and gas heaters.
1 mole CH4 = g
The molar mass of methane can be written as conversion factors.
16.0 g CH and mole CH4
1 mole CH g CH4

77. Learning Check
Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid.

78. Solution
Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass factors for acetic acid.
1 mole of acetic acid = g acetic acid
1 mole acetic acid and g acetic acid
60.0 g acetic acid mole acetic acid

79. Calculations with Molar Mass
Mole factors are used to convert between the grams of a substance and the number of moles.
Mole factor
Grams
Moles

80. 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 g Al = g Al
1 mole Al
mole factor for Al

81. 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?

82. 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)
= mole aspartame

83. Conservation of Mass
In a chemical reaction, the mass of the reactants is equal to the mass of the products moles Ag mole S = 1 mole Ag2S 2 (107.9 g) (32.1 g) = 1 (247.9 g)
247.9 g reactants = g product

84. Moles in Equations
We can read the equation in “moles” by placing the word “moles” between each coefficient and formula. 4 Fe O Fe2O3 4 moles Fe moles O moles Fe2O3

85. Writing Mole-Mole Factors
A mole-mole factor is a ratio of the coefficients for two substances.
4 Fe O Fe2O3
Fe and O mole Fe and 3 mole O2
3 mole O mole Fe
Fe and Fe2O mole Fe and 2 mole Fe2O3
2 mole Fe2O mole Fe
O2 and Fe2O mole O and 2 mole Fe2O3
2 mole Fe2O mole O2

86. Learning Check Consider the following equation: 3 H2 + N2 2 NH3
A. A mole factor for H2 and N2 is
1) 3 mole N ) 1 mole N ) 1 mole N2
1 mole H mole H mole H2
B. A mole factor for NH3 and H2 is
1) 1 mole H ) 2 mole NH ) 3 mole N2
2 mole NH mole H mole NH3

87. 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

88. Calculations with Mole Factors
Consider the following reaction:
4 Fe O 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

89. Learning Check Consider the following reaction: 4 Fe + 3 O2 2 Fe2O3
How many moles of Fe are needed to react with 12.0 moles of O2?
1) moles Fe
2) moles Fe
3) moles Fe

90. Solution 3) 16.0 moles Fe Consider the following reaction:
4 Fe O Fe2O3
How many moles of Fe are needed to react with 12.0 moles of O2?
12.0 mole O2 x 4 mole Fe = moles Fe
3 mole O2

91. Mass Calculations

92. Learning Check How many grams of O2 are needed to produce
0.400 mole of Fe2O3?
4 Fe O Fe2O3
1) g O2
2) g O2
3) g O2

93. Solution 2) 19.2 g O2 0.400 mole Fe2O3 x 3 mole O2 x 32.0 g O2
2 mole Fe2O3 1 mole O2
mole factor molar mass
= g O2

94. Calculating the Mass of a Reactant
The reaction between H2 and O2 produces 13.1 g of
water. How many grams of O2 reacted?
2H O2 2H2O
? g g
Plan: g H2O mole H2O mole O g O2
13.1 g H2O x 1 mole H2O x 1 mole O2 x g O g H2O mole H2O mole O2
= g O2

95. Learning Check 2 C2H2 + 5 O2 4 CO2 + 2 H2O
Acetylene gas C2H2 burns in the oxyactylene torch for welding. How many grams of C2H2 are burned if the reaction produces 75.0 g of CO2?
2 C2H O CO H2O
1) g C2H2
2) g C2H2
3) g C2H2

96. Solution 3) 22.2 g C2H2 2 C2H2 + 5 O2 4 CO2 + 2 H2O = 22.2 g C2H2
75.0 g CO2 x 1 mole CO2 x 2 moles C2H2 x g C2H2
44.0 g CO moles CO mole C2H2
= g C2H2

97. 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 cookies to eat
Percent yield 48 cookies x 100 = 80% yield cookies

98. 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 Theoretical Yield (g)

99. Sample Exercise % Yield
Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide.
2C O CO
What is the percent yield if 40.0 g of CO are produced from the reaction of 30.0 g O2?

100. Sample Exercise % Yield (cont.)
1. Calculate theoretical yield of CO.
30.0 g O2 x 1 mole O2 x 2 mole CO x g CO
32.0 g O mole O mole CO = g CO (theoretical)
2. Calculate the percent yield.
40.0 g CO (actual) x 100 = % yield
52.5 g CO(theoretical)

101. 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) NH3(g)
1) %
2) %
3) %

102. Solution 2) 56.5 % N2(g) + 3H2(g) 2NH3(g)
2) %
N2(g) + 3H2(g) NH3(g)
5.0 g H2 x 1 mole H2 x 2 moles NH3 x 17.0 g NH3
2.0 g H moles H mole NH3
= g NH3 (theoretical)
Percent yield = g NH3 x 100 = %
28.3 g NH3