3 Chemical Equations & Reaction Stoichiometry. 2 Chapter Three Goals 1.Chemical Equations...
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3 Chemical Equations & Reaction Stoichiometry. 2 Chapter Three Goals 1.Chemical Equations 2.Calculations Based on Chemical Equations 3.The Limiting Reactant
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
Slide 3
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
Slide 4
4 Chemical Equations Attempt to show on paper what is happening
at the laboratory and molecular levels.
Slide 5
5 Chemical Equations Look at the information an equation
provides:
Slide 6
6 Chemical Equations Look at the information an equation
provides: reactants yields products
Slide 7
7 Chemical Equations Look at the information an equation
provides: reactants yields products 1 formula unit 3 molecules 2
atoms 3molecules
Slide 8
8 Chemical Equations Look at the information an equation
provides: reactants yields products 1 formula unit 3 molecules 2
atoms 3 molecules 1 mole 3 moles 2 moles 3 moles
Slide 9
9 Chemical Equations Look at the information an equation
provides: 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 132g
Slide 10
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,C 3 H 8, burns in
oxygen to give carbon dioxide and water.
Slide 11
11 Law of Conservation of Matter NH 3 burns in oxygen to form
NO & water You do it!
Slide 12
12 Law of Conservation of Matter NH 3 burns in oxygen to form
NO & water
Slide 13
13 Law of Conservation of Matter C 7 H 16 burns in oxygen to
form carbon dioxide and water. You do it!
Slide 14
14 Law of Conservation of Matter C 7 H 16 burns in oxygen to
form carbon dioxide and water.
Slide 15
15 Law of Conservation of Matter C 7 H 16 burns in oxygen to
form carbon dioxide and water. Balancing equations is a skill
acquired only with lots of practice work many problems
Slide 16
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).
Slide 17
17 Calculations Based on Chemical Equations Example 3-1: How
many CO molecules are required to react with 25 formula units of Fe
2 O 3 ?
Slide 18
18 Calculations Based on Chemical Equations Example 3-2: How
many iron atoms can be produced by the reaction of 2.50 x 10 5
formula units of iron (III) oxide with excess carbon monoxide?
Slide 19
19 Calculations Based on Chemical Equations Example 3-2: How
many iron atoms can be produced by the reaction of 2.50 x 10 5
formula units of iron (III) oxide with excess carbon monoxide?
Slide 20
20 Calculations Based on Chemical Equations Example 3-2: How
many iron atoms can be produced by the reaction of 2.50 x 10 5
formula units of iron (III) oxide with excess carbon monoxide?
Slide 21
21 Calculations Based on Chemical Equations Example 3-3: What
mass of CO is required to react with 146 g of iron (III)
oxide?
Slide 22
22 Calculations Based on Chemical Equations Example 3-3: What
mass of CO is required to react with 146 g of iron (III)
oxide?
Slide 23
23 Calculations Based on Chemical Equations Example 3-3: What
mass of CO is required to react with 146 g of iron (III) oxide? YOU
MUST BE PROFICIENT WITH THESE TYPES OF PROBLEMS!!!
Slide 24
24 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?
Slide 25
25 Limiting Reactant Concept Mix PacketsEggsMilk 11 dozen1
gallon limiting reactant is the muffin mix 21 dozen1 gallon 31
dozen1 gallon 41 dozen1 gallon 51 dozen1 gallon 61 dozen1 gallon 71
dozen1 gallon limiting reactant is the dozen eggs
Slide 26
26 Limiting Reactant Concept Look at a chemical limiting
reactant situation. Zn + 2 HCl ZnCl 2 + H 2
Slide 27
27 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?
Slide 28
28 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?
Slide 29
29 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?
Slide 30
30 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?
Slide 31
31 Limiting Reactant Concept What do we do next? You do
it!
Slide 32
32 Limiting Reactant Concept Which is limiting reactant?
Limiting reactant is O 2. What is maximum mass of sulfur dioxide?
Maximum mass is 147 g.
Slide 33
33 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.
Slide 34
34 Percent Yields from Reactions Example 3-9: A 10.0 g sample
of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3
COOH, to produce 14.8 g of ethyl acetate, CH 3 COOC 2 H 5. What is
the percent yield?
Slide 35
35 Percent Yields from Reactions
Slide 36
36 Percent Yields from Reactions
Slide 37
37 Percent Yields from Reactions
Slide 38
38 Percent Yields from Reactions
Slide 39
39 Sequential Reactions Example 3-10: Starting with 10.0 g of
benzene (C 6 H 6 ), calculate the theoretical yield of nitrobenzene
(C 6 H 5 NO 2 ) and of aniline (C 6 H 5 NH 2 ).
Slide 40
40 Sequential Reactions
Slide 41
41 Sequential Reactions Next calculate the mass of aniline
produced. You do it!
Slide 42
42 Sequential Reactions
Slide 43
43 Sequential Reactions
Slide 44
44 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:
Slide 45
45 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?
Slide 46
46 Concentration of Solutions Example 3-12: Calculate the mass
of 8.00% w/w NaOH solution that contains 32.0 g of NaOH.
Slide 47
47 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!
Slide 48
48 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!
Slide 49
49 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.
Slide 50
50 Concentrations of Solutions Second common unit of
concentration:
Slide 51
51 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!
Slide 52
52 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.
Slide 53
53 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.
Slide 54
54 Concentrations of Solutions Example 3-16: Determine the mass
of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO 3 )
2. You do it!
Slide 55
55 Concentrations of Solutions Example 3-16: Determine the mass
of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO 3 )
2.
Slide 56
56 Concentrations of Solutions One of the reasons that molarity
is commonly used is because: M x L = moles solute and M x mL = mmol
solute M V = mol
Slide 57
57 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?
Slide 58
58 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?
Slide 59
59 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?
Slide 60
60 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 M 1 V 1 = M
2 V 2 is appropriate for dilutions, but not for chemical
reactions.
Slide 61
61 Dilution of Solutions Common method to dilute a solution
involves the use of volumetric flask, pipet, and suction bulb.
Slide 62
62 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?
Slide 63
63 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!
Slide 64
64 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?
Slide 65
65 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.
Slide 66
66 Using Solutions in Chemical Reactions Example 3-20: What
volume of 0.500 M BaCl 2 is required to completely react with 4.32
g of Na 2 SO 4 ?
Slide 67
67 Using Solutions in Chemical Reactions Example 3-20: What
volume of 0.500 M BaCl 2 is required to completely react with 4.32
g of Na 2 SO 4 ?
Slide 68
68 Using Solutions in Chemical Reactions Example 3-20: What
volume of 0.500 M BaCl 2 is required to completely react with 4.32
g of Na 2 SO 4 ?
Slide 69
69 Using Solutions in Chemical Reactions Example 3-21: (a)What
volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum
nitrate, Al(NO 3 ) 3 ?
Slide 70
70 Using Solutions in Chemical Reactions Example 3-20: (a)What
volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum
nitrate?
Slide 71
71 Using Solutions in Chemical Reactions (b)What mass of Al(OH)
3 precipitates in (a)? You do it!
Slide 72
72 Using Solutions in Chemical Reactions (b) What mass of
Al(OH) 3 precipitates in (a)?
Slide 73
73 Using Solutions in Chemical Reactions Titrations are 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 indicator also
Slide 74
74 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?
Slide 75
75 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?
Slide 76
76 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?
Slide 77
77 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?