TEMA PARA LA MATERIA DE QUÍMICA...t/t

1.  CHEMICAL EQUATIONS & REACTION
STOICHIOMETRY
PROFESOR: QBA MIGUEL ÁNGEL CASTRO RAMÍREZ
1

2. Chapter Three Goals
1. Chemical Equations
2. Calculations Based on Chemical Equations
3. The Limiting Reactant Concept
4. Percent Yields from Chemical Reactions
5. Sequential Reactions
6. Concentrations of Solutions
7. Dilution of solutions
8. Using Solutions in Chemical Reactions
9. Synthesis Question
2

3. Chemical Equations
 Symbolic representation of a chemical reaction
that shows:
1. reactants on left side of reaction
2. products on right side of equation
3. relative amounts of each using stoichiometric
coefficients
3

4. Chemical Equations
 Attempt to show on paper what is happening at
the laboratory and molecular levels.
4

5. Chemical Equations
 Look at the information an equation provides:
∆
Fe 2 O 3 + 3 CO  → 2 Fe + 3 CO 2

5

6. Chemical Equations
 Look at the information an equation provides:
∆
Fe 2 O 3 + 3 CO  → 2 Fe + 3 CO 2

reactants yields products
6

7. Chemical Equations
 Look at the information an equation provides:
∆
Fe 2 O 3 + 3 CO  → 2 Fe + 3 CO 2

reactants yields products
1 formula unit 3 molecules 2 atoms 3 molecules
7

8. Chemical Equations
 Look at the information an equation provides:
∆
Fe 2 O 3 + 3 CO  → 2 Fe + 3 CO 2

reactants yields products
1 formula unit 3 molecules 2 atoms 3 molecules
1 mole 3 moles 2 moles 3 moles
8

9. Chemical Equations
 Look at the information an equation provides:
∆
Fe 2 O 3 + 3 CO  → 2 Fe + 3 CO 2

reactants yields products
1 formula unit 3 molecules 2 atoms 3 molecules
1 mole 3 moles 2 moles 3 moles
159.7 g 84.0 g 111.7 g 132 g
9

10. Chemical Equations
 Law of Conservation of Matter
– There is no detectable change in quantity of matter in an
ordinary chemical reaction.
– Balanced chemical equations must always include the same
number of each kind of atom on both sides of the equation.
– This law was determined by Antoine Lavoisier.
 Propane,C3H8, burns in oxygen to give carbon dioxide
and water.
∆
C H +5O 
3 8 → 3 CO + 4 H O
2 2 2
10

11. Law of Conservation of Matter
 NH3 burns in oxygen to form NO & water
You do it!
11

12. Law of Conservation of Matter
 NH3 burns in oxygen to form NO & water
∆
2 NH 3 + O 2 
→ 2 NO + 3 H 2 O
5
2
or correctly
∆
4 NH 3 + 5 O 2 
→ 4 NO + 6 H 2 O
12

13. Law of Conservation of Matter
 C7H16
burns in oxygen to form carbon dioxide
and water.
You do it!
13

14. Law of Conservation of Matter
 C7H16
burns in oxygen to form carbon dioxide
and water.
∆
C 7 H16 + 11 O 2 → 7 CO 2 + 8 H 2 O
14

15. Law of Conservation of Matter
 C7H16
burns in oxygen to form carbon dioxide
and water.
∆
C 7 H16 + 11 O 2 → 7 CO 2 + 8 H 2 O
 Balancing equations is a skill acquired only
with lots of practice
– work many problems
15

16. Calculations Based on Chemical
Equations
 Can work in moles, formula units, etc.
 Frequently, we work in mass or weight (grams
or kg or pounds or tons).
∆
Fe 2 O 3 + 3 CO  → 2 Fe + 3 CO 2

16

17. Calculations Based on Chemical
Equations
 Example 3-1: How many CO molecules are
required to react with 25 formula units of
Fe2O3?
3 CO molecules
? CO molecules = 25 formula units Fe 2 O 3 ×
1 Fe 2 O 3 formula unit
= 75 molecules of CO
17

18. Calculations Based on Chemical
Equations
 Example 3-2: How many iron atoms can be
produced by the reaction of 2.50 x 105 formula
units of iron (III) oxide with excess carbon
monoxide?
? Fe atoms = 2.50 × 10 formula units Fe 2 O 3
5
18

19. Calculations Based on Chemical
Equations
 Example 3-2: How many iron atoms can be
produced by the reaction of 2.50 x 105 formula
units of iron (III) oxide with excess carbon
monoxide?
? Fe atoms = 2.50 × 10 formula units Fe 2 O 3
5
2 Fe atoms
× =
1 formula units Fe 2 O 3
19

20. Calculations Based on Chemical
Equations
 Example 3-2: How many iron atoms can be
produced by the reaction of 2.50 x 105 formula
units of iron (III) oxide with excess carbon
monoxide?
? Fe atoms = 2.50 × 10 formula units Fe 2 O 3
5
2 Fe atoms
× = 5.00 × 10 Fe atoms
5
1 formula units Fe 2 O 3
20

21. Calculations Based on Chemical
Equations
 Example 3-3: What mass of CO is required to
react with 146 g of iron (III) oxide?
1 mol Fe 2 O 3
? g CO = 146 g Fe 2 O 3 ×
159.7 g Fe 2 O 3
21

22. Calculations Based on Chemical
Equations
 Example 3-3: What mass of CO is required to
react with 146 g of iron (III) oxide?
1 mol Fe 2 O 3 3 mol CO
? g CO = 146 g Fe 2O 3 × ×
159.7 g Fe 2O 3 1 mol Fe 2O 3
22

23. Calculations Based on Chemical
Equations
 Example 3-3: What mass of CO is required to
react with 146 g of iron (III) oxide?
1 mol Fe 2 O 3 3 mol CO
? g CO = 146 g Fe 2O 3 × ×
159.7 g Fe 2O 3 1 mol Fe 2O 3
28.0 g CO
× = 76.8 g CO
1 mol CO
23

24. Calculations Based on Chemical
Equations
 Example 3-4: What mass of carbon dioxide can
be produced by the reaction of 0.540 mole of
iron (III) oxide with excess carbon monoxide?
3 mol CO 2
? g CO 2 = 0.540 mol Fe 2 O3 ×
1 mol Fe 2 O3
24

25. Calculations Based on Chemical
Equations
 Example 3-4: What mass of carbon dioxide can
be produced by the reaction of 0.540 mole of
iron (III) oxide with excess carbon monoxide?
3 mol CO 2 44.0 g CO 2
? g CO 2 = 0.540 mol Fe 2 O 3 × ×
1 mol Fe 2 O 3 1 mol CO 2
25

26. Calculations Based on Chemical
Equations
 Example 3-4: What mass of carbon dioxide can
be produced by the reaction of 0.540 mole of
iron (III) oxide with excess carbon monoxide?
3 mol CO 2 44.0 g CO 2
? g CO 2 = 0.540 mol Fe 2 O 3 × ×
1 mol Fe 2 O 3 1 mol CO 2
= 71.3 g CO 2
26

27. Calculations Based on Chemical
Equations
 Example 3-5: What mass of iron (III) oxide
reacted with excess carbon monoxide if the
carbon dioxide produced by the reaction had a
mass of 8.65 grams?
You do it!
27

28. Calculations Based on Chemical
Equations
 Example 3-5: What mass of iron (III) oxide
reacted with excess carbon monoxide if the
carbon dioxide produced by the reaction had a
mass of 8.65 grams?
1 molCO2 1mol Fe2O3
? g Fe2O3 = 8.65 g CO2 × × ×
44.0 g CO2 3 mol CO2
159.7 g Fe2O3
= 10.5 g Fe2O3
1 mol Fe2O3
28

29. Calculations Based on Chemical
Equations
 Example 3-6: How many pounds of carbon
monoxide would react with 125 pounds of iron
(III) oxide?
You do it!
29

30. Calculations Based on Chemical
Equations
454 g Fe 2 O 3
? lb CO = 125 lb Fe 2 O 3 ×
1 lb Fe 2 O 3
1 mol Fe 2 O 3 3 mol CO
× × ×
159.7 g Fe 2 O 3 1 mol Fe 2 O 3
28 g CO 1 lb CO
× = 65.7 lb CO
1 mol CO 454 g CO
YOU MUST BE PROFICIENT WITH THESE
TYPES OF PROBLEMS!!!
Now go to your text and work the problems assigned!
30

31. Limiting Reactant Concept
 Kitchen example of limiting reactant concept.
1 packet of muffin mix + 2 eggs + 1 cup of milk
→ 12 muffins
 How many muffins can we make with the
following amounts of mix, eggs, and milk?
31

32. Limiting Reactant Concept
 Mix Packets Eggs Milk
1 1 dozen 1 gallon
limiting reactant is the muffin mix
2 1 dozen 1 gallon
3 1 dozen 1 gallon
4 1 dozen 1 gallon
5 1 dozen 1 gallon
6 1 dozen 1 gallon
7 1 dozen 1 gallon
limiting reactant is the dozen eggs
32

33. Limiting Reactant Concept
 Example 3-7: Suppose a box contains 87 bolts, 110
washers, and 99 nuts. How many sets, each consisting
of one bolt, two washers, and one nut, can you construct
from the contents of one box?
(
87 bolts 1 set )
1 bolt
= 87 sets
(
110 washers 1 set )
2 washers
= 55 sets
(
99 nuts 1 set )
1 nut
= 99 sets
the maximum number we can make is 55 sets
determined by the smallest number
33

34. Limiting Reactant Concept
 Look at a chemical limiting reactant situation.
Zn + 2 HCl→ ZnCl2 + H2
34

35. Limiting Reactant Concept
 Example 3-8: What is the maximum mass of sulfur
dioxide that can be produced by the reaction of 95.6 g
of carbon disulfide with 110. g of oxygen?
CS2 + 3 O 2 → CO 2 + 2 SO 2
35

36. Limiting Reactant Concept
 Example 3-8: What is the maximum mass of sulfur
dioxide that can be produced by the reaction of 95.6 g
of carbon disulfide with 110. g of oxygen?
CS2 + 3 O 2 → CO 2 + 2 SO 2
1 mol 3 mol 1 mol 2 mol
36

37. Limiting Reactant Concept
 Example 3-8: What is the maximum mass of sulfur
dioxide that can be produced by the reaction of 95.6 g
of carbon disulfide with 110. g of oxygen?
CS2 + 3 O 2 → CO 2 + 2 SO 2
1 mol 3 mol 1 mol 2 mol
76.2 g 3(32.0 g) 44.0 g 2(64.1 g)
37

38. Limiting Reactant Concept
 Example 3-8: What is the maximum mass of sulfur
dioxide that can be produced by the reaction of 95.6 g
of carbon disulfide with 110. g of oxygen?
CS2 + 3 O 2 → CO 2 + 2 SO 2
1 mol CS2
? mol SO 2 = 95.6 g CS2 ×
76.2 g
38

39. Limiting Reactant Concept
CS2 + 3 O 2 → CO 2 + 2 SO 2
1 mol CS 2
? mol SO 2 = 95.6 g CS2 ×
76.2 g
2 mol SO 2 64.1 g SO 2
× × =161 g SO 2
1 mol CS2 1 mol SO 2
What do we do next?
You do it!
39

40. Limiting Reactant Concept
CS2 + 3 O 2 → CO 2 + 2 SO 2
1 mol CS2 2 mol SO 2 64.1 g SO 2
? mol SO 2 = 95.6 g CS2 × × × = 161 g SO 2
76.2 g 1 mol CS2 1 mol SO 2
1 mol O 2 2 mol SO 2 64.1 g SO 2
? mol SO 2 = 110 g O 2 × × × = 147 g SO 2
32.0 g O 2 3 mol O 2 1 mol SO 2
 Which is limiting reactant?
 Limiting reactant is O2.
 What is maximum mass of sulfur dioxide?
 Maximum mass is 147 g.
40

41. Percent Yields from Reactions
 Theoretical yield is calculated by assuming that the
reaction goes to completion.
– Determined from the limiting reactant calculation.
 Actual yield is the amount of a specified pure product
made in a given reaction.
– In the laboratory, this is the amount of product that is formed in
your beaker, after it is purified and dried.
 Percent yield indicates how much of the product is
obtained from a reaction.
actual yield
% yield = × 100%
theoretical yield
41

42. Percent Yields from Reactions
 Example3-9: A 10.0 g sample of ethanol,
C2H5OH, was boiled with excess acetic acid,
CH3COOH, to produce 14.8 g of ethyl acetate,
CH3COOC2H5. What is the percent yield?
42

43. Percent Yields from Reactions
CH 3COOH + C 2 H 5OH → CH 3COOC 2 H 5 + H 2 O
1. Calculate the theoretical yield
43

44. Percent Yields from Reactions
CH 3COOH + C 2 H 5OH → CH 3COOC 2 H 5 + H 2 O
1. Calculate the theoretical yield
88.0 g CH 3COOC 2 H 5
? g CH 3COOC 2 H 5 = 10.0 g C 2 H 5OH ×
46.0 g C 2 H 5OH
= 19.1 g CH 3COOC 2 H 5
44

45. Percent Yields from Reactions
CH 3COOH + C 2 H 5OH → CH 3COOC 2 H 5 + H 2 O
1. Calculate the theoretical yield
88.0 g CH 3COOC 2 H 5
? g CH 3COOC 2 H 5 = 10.0 g C 2 H 5OH ×
46.0 g C 2 H 5OH
= 19.1 g CH 3COOC 2 H 5
2. Calculate the percent yield.
45

46. Percent Yields from Reactions
CH 3COOH + C 2 H 5OH → CH 3COOC 2 H 5 + H 2 O
1. Calculate the theoretical yield
88.0 g CH 3COOC 2 H 5
? g CH 3COOC 2 H 5 = 10.0 g C 2 H 5OH ×
46.0 g C 2 H 5OH
= 19.1 g CH 3COOC 2 H 5
2. Calculate the percent yield.
14.8 g CH 3COOC 2 H 5
% yield = × 100% = 77.5%
19.1 g CH 3COOC 2 H 5
46

47. Sequential Reactions
N O2 NH2
HNO3 Sn
H2SO4 Conc. HCl
benzene nitrobenzene aniline
 Example 3-10: Starting with 10.0 g of benzene (C6H6),
calculate the theoretical yield of nitrobenzene (C6H5NO2)
and of aniline (C6H5NH2).
47

48. Sequential Reactions
1 mol benzene
? g nitrobenzene = 10.0 g benzene × ×
78.0 g benzene
48

49. Sequential Reactions
1 mol benzene
? g nitrobenzene = 10.0 g benzene × ×
78.0 g benzene
1 mol nitrobenzene 123.0 g nitrobenzene
× = 15.8 g nitrobenzene
1 mol benzene 1 mol nitrobenzene
 Next calculate the mass of aniline produced.
You do it!
49

50. Sequential Reactions
N O2 NH2
HNO3 Sn
H2SO4 Conc. HCl
benzene nitrobenzene aniline
1 mol nitrobenzene
? g aniline = 15.8 g nitrobenzene × ×
123.0 g nitrobenzene
50

51. Sequential Reactions
N O2 NH2
HNO3 Sn
H2SO4 Conc. HCl
benzene nitrobenzene aniline
1 mol nitrobenzene
? g aniline = 15.8 g nitrobenzene × ×
123.0 g nitrobenzene
1 mol aniline 93.0 g aniline
× = 11.9 g aniline
1 mol nitrobenzene 1 mol aniline
51

52. Sequential Reactions
 If6.7 g of aniline is prepared from 10.0 g of
benzene, what is the percentage yield?
You do it!
6.7 g aniline
% yield = ×100% = 56%
11.9 g aniline
52

53. Concentration of Solutions
 Solution is a mixture of two or more substances dissolved in
another.
– Solute is the substance present in the smaller amount.
– Solvent is the substance present in the larger amount.
– In aqueous solutions, the solvent is water.
 The concentration of a solution defines the amount of solute
dissolved in the solvent.
– The amount of sugar in sweet tea can be defined by its concentration.
 One common unit of concentration is:
mass of solute
% by mass of solute = ×100%
mass of solution
mass of solution = mass of solute + mass of solvent
53 % by mass of solute has the symbol % w/w

54. Concentration of Solutions
 Example
3-12: Calculate the mass of 8.00%
w/w NaOH solution that contains 32.0 g of
NaOH.
100.0 g solution
? g solution = 32.0 g NaOH ×
8.00 g NaOH
= 400. g sol' n
54

55. Concentration of Solutions
 Example 3-11: What mass of NaOH is required to
prepare 250.0 g of solution that is 8.00% w/w NaOH?
 8.00 g NaOH 
250.0 g solution
 100.0 g solution  = 20.0 g NaOH

 
55

56. Concentration of Solutions
 Example 3-13: Calculate the mass of NaOH in 300.0
mL of an 8.00% w/w NaOH solution. Density is 1.09
g/mL.
You do it!
1.09 g sol' n
? g NaOH = 300.0 mL sol' n × ×
1 mL sol' n
8.00 g NaOH
= 26.2 g NaOH
100 g sol' n
56

57. Concentrations of Solutions
 Example 3-14: What volume of 12.0% KOH
contains 40.0 g of KOH? The density of the
solution is 1.11 g/mL.
You do it!
57

58. Concentrations of Solutions
 Example 3-14: What volume of 12.0% KOH
contains 40.0 g of KOH? The density of the
solution is 1.11 g/mL.
100.0 g solution 1 mL solution
? mL solution = 40.0 g KOH × ×
12.0 g KOH 1.11 g solution
= 300. mL solution
58

59. Concentrations of Solutions
 Second common unit of concentration:
number of moles of solute
molarity =
number of liters of solution
moles
M=
L
mmol
M=
mL
59

60. Concentrations of Solutions
 Example 3-15: Calculate the molarity of a solution that
contains 12.5 g of sulfuric acid in 1.75 L of solution.
You do it!
60

61. Concentrations of Solutions
 Example 3-15: Calculate the molarity of a
solution that contains 12.5 g of sulfuric acid in
1.75 L of solution.
? mol H 2SO 4 12.5 g H 2SO 4 1 mol H 2SO 4
= ×
L sol' n 1.75 L sol' n 98.1 g H 2SO 4
61

62. Concentrations of Solutions
 Example 3-15: Calculate the molarity of a
solution that contains 12.5 g of sulfuric acid in
1.75 L of solution.
? mol H 2SO 4 12.5 g H 2SO 4 1 mol H 2SO 4
= ×
L sol' n 1.75 L sol' n 98.1 g H 2SO 4
0.0728 mol H 2SO 4
=
L
= 0.0728 M H 2SO 4
62

63. Concentrations of Solutions
 Example 3-16: Determine the mass of calcium
nitrate required to prepare 3.50 L of 0.800 M
Ca(NO3)2 .
You do it!
63

64. Concentrations of Solutions
 Example 3-16: Determine the mass of calcium
nitrate required to prepare 3.50 L of 0.800 M
Ca(NO3)2 .
0.800 mol Ca(NO 3 ) 2
? g Ca(NO 3 ) 2 = 3.50 L × ×
L
164 g Ca(NO 3 ) 2
= 459 g Ca(NO 3 ) 2
1 mol Ca(NO 3 ) 2
64

65. Concentrations of Solutions
 Oneof the reasons that molarity is commonly
used is because:
M x L = moles solute
and
M x mL = mmol solute
65

66. Concentrations of Solutions
 Example 3-17: The specific gravity of concentrated
HCl is 1.185 and it is 36.31% w/w HCl. What is its
molarity?
specific gravity = 1.185 tells us
density =1.185 g/mL or 1185 g/L
66

67. Concentrations of Solutions
 Example 3-17: The specific gravity of concentrated
HCl is 1.185 and it is 36.31% w/w HCl. What is its
molarity?
specific gravity = 1.185 tells us
density =1.185 g/mL or 1185g/L
1185 g solution 36.31 g HCl
? mol HCl/L = × ×
L solution 100 g sol' n
67

68. Concentrations of Solutions
 Example 3-17: The specific gravity of concentrated
HCl is 1.185 and it is 36.31% w/w HCl. What is its
molarity?
specific gravity = 1.185 tells us
density =1.185 g/mL or 1185g/L
1185 g solution 36.31 g HCl
? mol HCl/L = × ×
L solution 100 g sol' n
1 mol HCl
= 11.80 M HCl
36.46 g HCl
68

69. Dilution of Solutions
 To dilute a solution, add solvent to a
concentrated solution.
– One method to make tea “less sweet.”
– How fountain drinks are made from syrup.
 The number of moles of solute in the two
solutions remains constant.
 The relationship M1V1 = M2V2 is appropriate for
dilutions, but not for chemical reactions.
69

70. Dilution of Solutions
 Common method to dilute a solution involves
the use of volumetric flask, pipet, and suction
bulb.
70

71. Dilution of Solutions
 Example 3-18: If 10.0 mL of 12.0 M HCl is
added to enough water to give 100. mL of
solution, what is the concentration of the
solution?
M 1V1 = M 2 V2
12.0 M ×10.0 mL = M 2 ×100.0 mL
12.0 M ×10.0 mL
M2 =
100.0 mL
=1.20 M
71

72. Dilution of Solutions
 Example 3-19: What volume of 18.0 M sulfuric
acid is required to make 2.50 L of a 2.40 M
sulfuric acid solution?
You do it!
72

73. Dilution of Solutions
 Example 3-19: What volume of 18.0 M sulfuric
acid is required to make 2.50 L of a 2.40 M
sulfuric acid solution?
M 1 V1 = M 2 V2
M 2 × V2
V1 =
M1
2.50 L × 2.40 M
V1 =
18.0 M
73
= 0.333 L or 333 mL

74. Using Solutions in Chemical
Reactions
 Combine the concepts of molarity and
stoichiometry to determine the amounts of
reactants and products involved in reactions in
solution.
74

75. Using Solutions in Chemical
Reactions
 Example 3-20: What volume of 0.500 M BaCl2 is
required to completely react with 4.32 g of Na2SO4?
Na 2SO 4 + BaCl 2 → BaSO 4 + 2 NaCl
75

76. Using Solutions in Chemical
Reactions
 Example 3-20: What volume of 0.500 M BaCl2 is
required to completely react with 4.32 g of Na2SO4?
Na 2SO 4 + BaCl 2 → BaSO 4 + 2 NaCl
1 mol Na 2SO 4
? L BaCl 2 = 4.32 gNa 2SO 4 × ×
142 g Na 2SO 4
76

77. Using Solutions in Chemical
Reactions
 Example 3-20: What volume of 0.500 M BaCl2 is
required to completely react with 4.32 g of Na2SO4?
Na 2SO 4 + BaCl 2 → BaSO 4 + 2 NaCl
1 mol Na 2SO 4
? L BaCl 2 = 4.32 gNa 2SO 4 × ×
142 g Na 2SO 4
1 mol BaCl 2 1 L BaCl 2
× = 0.0608 L
1 mol Na 2SO 4 0.500 mol BaCl 2
77

78. Using Solutions in Chemical
Reactions
 Example3-21: (a)What volume of 0.200 M
NaOH will react with 50.0 mL 0f 0.200 M
aluminum nitrate, Al(NO3)3?
Al ( NO3 ) 3 + 3 NaOH → Al ( OH ) 3 + 3 NaNO3
You do it!
78

79. Using Solutions in Chemical
Reactions
 Example3-20: (a)What volume of 0.200 M
NaOH will react with 50.0 mL 0f 0.200 M
aluminum nitrate?
Al( NO 3 ) 3 + 3 NaOH →Al(OH)3 + 3 NaNO3
1L
? mL NaOH = 50.0 mL Al(NO3 ) 3 sol' n ×
1000 mL
0.200 mol Al(NO3 ) 3 sol' n 3 mol NaOH
× ×
1 L Al(NO3 ) 3 sol' n 1 mol Al(NO3 ) 3
1 L NaOH
= 0.150 L or 150 mL NaOH sol' n
79 0.200 mol NaOH

80. Using Solutions in Chemical
Reactions
 (b)What mass of Al(OH)3 precipitates in (a)?
You do it!
80

81. Using Solutions in Chemical
Reactions
 (b) What mass of Al(OH)3 precipitates in (a)?
1L
? g Al(OH) 3 = 50.0 mL Al(NO 3 )3 sol' n ×
1000 mL
0.200 mol Al(NO 3 )3 1 mol Al(OH) 3 78.0 g Al(OH) 3
× ×
1 L Al(NO 3 )3 sol' n 1 mol Al(NO 3 )3 1 mol Al(OH) 3
= 0.780 g Al(OH) 3
81

82. Using Solutions in Chemical
Reactions
 Titrationsare a method of determining the
concentration of an unknown solutions from the
known concentration of a solution and solution
reaction stoichiometry.
– Requires special lab glassware
 Buret, pipet, and flasks
– Must have an an indicator also
82

83. Using Solutions in Chemical
Reactions
 Example 3-22: What is the molarity of a KOH solution
if 38.7 mL of the KOH solution is required to react with
43.2 mL of 0.223 M HCl?
KOH + HCl → KCl + H 2 O
83

84. Using Solutions in Chemical
Reactions
 Example 3-22: What is the molarity of a KOH solution
if 38.7 mL of the KOH solution is required to react with
43.2 mL of 0.223 M HCl?
KOH + HCl → KCl + H 2 O
43.2 mL × 0.223 M HCl = 9.63 mmol HCl
84

85. Using Solutions in Chemical
Reactions
 Example 3-22: What is the molarity of a KOH solution
if 38.7 mL of the KOH solution is required to react with
43.2 mL of 0.223 M HCl?
KOH + HCl → KCl + H 2 O
43.2 mL × 0.223 M HCl = 9.63 mmol HCl
1 mmol KOH
9.63 mmol HCl × = 9.63 mmol KOH
1 mmol HCl
85

86. Using Solutions in Chemical
Reactions
 Example 3-22: What is the molarity of a KOH solution
if 38.7 mL of the KOH solution is required to react with
43.2 mL of 0.223 M HCl?
KOH + HCl → KCl + H 2 O
43.2 mL × 0.223 M HCl = 9.63 mmol HCl
1 mmol KOH
9.63 mmol HCl × = 9.63 mmol KOH
1 mmol HCl
9.63 mmol KOH
= 0.249 M KOH
38.7 mL KOH
86

87. Using Solutions in Chemical
Reactions
 Example 3-23: What is the molarity of a barium
hydroxide solution if 44.1 mL of 0.103 M HCl is
required to react with 38.3 mL of the Ba(OH) 2 solution?
Ba(OH)2 + 2 HCl → BaCl 2 + 2 H 2 O
(44.1 mL HCl)(0.103 M HCl) = 4.54 mmol HCl
87

88. Using Solutions in Chemical
Reactions
 Example 3-23: What is the molarity of a barium
hydroxide solution if 44.1 mL of 0.103 M HCl is
required to react with 38.3 mL of the Ba(OH) 2 solution?
Ba(OH)2 + 2 HCl → BaCl 2 + 2 H 2 O
(44.1 mL HCl)(0.103 M HCl) = 4.54 mmol HCl
1 mmol Ba(OH)2
4.54 mmol HCl ×
2 mmol HCl
88

89. Using Solutions in Chemical
Reactions
 Example 3-23: What is the molarity of a barium
hydroxide solution if 44.1 mL of 0.103 M HCl is
required to react with 38.3 mL of the Ba(OH) 2 solution?
Ba(OH)2 + 2 HCl → BaCl 2 + 2 H 2 O
(44.1 mL HCl)(0.103 M HCl) = 4.54 mmol HCl
1 mmol Ba(OH)2
4.54 mmol HCl ×
2 mmol HCl
= 2.27 mmol Ba(OH)2
89

90. Using Solutions in Chemical
Reactions
 Example 3-23: What is the molarity of a barium
hydroxide solution if 44.1 mL of 0.103 M HCl is
required to react with 38.3 mL of the Ba(OH) 2 solution?
Ba(OH)2 + 2 HCl → BaCl 2 + 2 H 2 O
(44.1 mL HCl)(0.103 M HCl) = 4.54 mmol HCl
1 mmol Ba(OH)2
4.54 mmol HCl ×
2 mmol HCl
= 2.27 mmol Ba(OH)2
2.27 mL Ba(OH)2
= 0.0593M Ba(OH)2
90 38.3 mL Ba(OH) 2

91. Synthesis Question
 Nylonis made by the reaction of
hexamethylene diamine
CH2 CH2 CH2 NH2
H2N CH2 CH2 CH2
with adipic acid.
HO H2 H2
C O
C C C
C C
O H2 H2 OH
91

92. Synthesis Question
in a 1 to 1 mole ratio. The structure of nylon is:
H2 H2 O H2 H2 H2
C C C C C C
* C C C N C C N n *
H2 H2 H H2 H2 H
O
where the value of n is typically 450,000. On a daily
basis, a DuPont factory makes 1.5 million pounds of
nylon. How many pounds of hexamethylene diamine
and adipic acid must they have available in the plant
each day?
92

93. Synthesis Question
Molar mass of 1 nylon molecule = [(12 × 12) + (1× 22) + ( 2 × 14) + (2 × 16)] 450,000
       
   
C atoms H atoms N atoms O atoms # of units
= [226 g/mol] 450,000
= 1.02 × 108 g/mol nylon molecules
93

94. Synthesis Question
Molar mass of 1 nylon molecule = [(12 × 12) + (1× 22) + ( 2 × 14) + (2 × 16)] 450,000
       
   
C atoms H atoms N atoms O atoms # of units
= [226 g/mol] 450,000
= 1.02 × 108 g/mol nylon molecules
 454 g 
1.5 million pounds = (1.5 × 106 lb) 
 lb 
= 6.81× 108 g
94

95. Synthesis Question
Molar mass of 1 nylon molecule = [(12 × 12) + (1× 22) + ( 2 ×14) + (2 ×16)] 450,000
       
   
C atoms H atoms N atoms O atoms # of units
= [226 g/mol] 450,000
= 1.02 ×108 g/mol nylon molecules
 454 g 
1.5 million pounds = (1.5 ×10 lb) 6

 lb 
= 6.81×108 g
 1 mol nylon 
(
# mol of nylon molecules = 6.81×108 g  )
 1.02 ×108 g 

 
= 6.68 mol of nylon
95

96. Synthesis Question
Because the nylon formation reaction uses 1 mole of adipic acid × 450,000
plus 1 mole of hexamethylene diamine × 450,000 per mole of nylon formed,
to make 6.68 mol of nylon requires :
96

97. Synthesis Question
Because the nylon formation reaction uses 1 mole of adipic acid × 450,000
plus 1 mole of hexamethylene diamine × 450,000 per mole of nylon formed,
to make 6.68 mol of nylon requires :
 1 lb 
( )
adipic acid - 6.68 × 450,000 ×146 g/mol = 4.39 ×108 g   = 9.66 ×105 lb
 454 g 
 
97

98. Synthesis Question
Because the nylon formation reaction uses 1 mole of adipic acid × 450,000
plus 1 mole of hexamethylene diamine × 450,000 per mole of nylon formed,
to make 6.68 mol of nylon requires :
 1 lb 
( )
adipic acid - 6.68 × 450,000 ×146 g/mol = 4.39 ×108 g 
 454 g  = 9.66 ×10 lb

5
 
 1 lb 
(
hexamethylene diamine - 6.68 × 450,000 ×116 g/mol = 3.49 ×108 g  ) 454 g  = 7.68 ×10 lb

5
 
98

99. Group Activity
Manganese dioxide, potassium hydroxide and
oxygen react in the following fashion:
4 MnO 2 + 4 KOH + 3 O 2 → 4 KMnO 4 + 2 H 2 O
A mixture of 272.9 g of MnO2, 26.6 L of 0.250
M KOH, and 41.92 g of O2 is allowed to react
as shown above. After the reaction is finished,
234.6 g of KMnO4 is separated from the
reaction mixture. What is the per cent yield of
99 this reaction?

100. End of Chapter 3
100