Download Chemical Reaction

Lecture 4 Chemical Reactions and
Quantities
Chemical Reactions and Equations
Copyright © 2007 by Pearson Education, Inc.
Publishing as Benjamin Cummings
1
Post-exam Error Analysis
 Work in groups of 3 on exam
 Maximum of 15 points possible
15 – (1/5)(theoretical – actual)
 This is what I call PEA score
 Exam Correction Factor
 (wrong answer points / right answer points)
 Points you earned: Correction factor x PEA score
 Examples on the board

2
Chemical Reaction
In a chemical reaction,
 A chemical change
produces one or more
new substances.
 There is a change in the
composition of one or
more substances.
 Old bonds are broken and
new ones are formed
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3
Chemical Equations
A chemical equation
 Gives the chemical formulas of the reactants on the
left of the arrow and the products on the right.
Reactants
O2 (g)
Product
CO2 (g)
C(s)
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4
Symbols Used in Equations
Symbols are used in
chemical equations
to show
TABLE 5.2
 The states of the
reactants.
 The states of the
products.
 The reaction conditions.
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5
Chemical Equations are Balanced
Chemical equations must
be balanced!
 Atoms are not gained
or lost.
 The number of atoms
in the reactants is
equal to the number of
atoms in the products.
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Inc. Publishing as Benjamin Cummings
6
A Balanced Chemical Equation
In a balanced chemical equation,
 There must be the same number of each type of
atom on the reactant side and on the product side.
Al +
S
2Al + 3S
Al2S3
Not Balanced
Al2S3
Balanced
2Al
=
2Al
3S
=
3S
7
Learning Check
State the number of atoms of each element on the
reactant side and the product side for each of the
following balanced equations:
A. P4(s) + 6Br2(l) → 4 PBr3(g)
B. 2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s)
8
Learning Check
Determine if each equation is balanced or not.
A. Na(s) + N2(g) → Na3N(s)
B. C2H4(g)
+ H2O(l) → C2H5OH(l)
9
Checking a Balanced Equation
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Reactants
1 C atom
4 H atoms
4 O atoms
=
=
=
Products
1 C atom
4 H atoms
4 O atoms
10
Guide to Balancing Equations
 Write out correct formulas
 Count the atoms of each element on both
sides of the equation
 Never change subscripts
 Use coefficients to balance each element
 Do oxygen last
 Hydrogen is next to last
 Check final equation for balance
11
Balancing Chemical Equations
Balance the following chemical equation:
NH3(g) + O2(g)  NO(g) + H2O(g)
12
Learning Check
Check the balance of atoms in the following:
Fe3O4(s) + 4H2(g)
3Fe(s) + 4H2O(l)
A. Number of H atoms in products.
1) 2
2) 4
3) 8
B. Number of O atoms in reactants.
1) 2
2) 4
3) 8
C. Number of Fe atoms in reactants.
1) 1
2) 3
3) 4
13
Learning Check
Balance each equation and list the coefficients in the
balanced equation going from reactants to products:
A. __Mg(s)
+ _N2(g)
1) 1, 3, 2
B. __Al(s)
1) 3, 3, 2
__Mg3N2(s)
2) 3, 1, 2
+ __Cl2(g)
3) 3, 1, 1
__AlCl3(s)
2) 1, 3, 1
3) 2, 3, 2
14
Equations with Polyatomic Ions
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15
Learning Check
Barium nitrate reacts with sodium carbonate to
form sodium nitrate and barium carbonate.
What is the balanced chemical equation?
16
Learning Check
Balance and list the coefficients from reactants to products:
A. __Fe2O3(s) + __C(s)
1) 2, 3, 2,3
__Fe(s)
2) 2, 3, 4, 3
3) 1, 1, 2, 3
B. __Al(s) + __FeO(s)
1) 2, 3, 3, 1
1) 3, 2, 1, 2
__Fe(s) +
2) 2, 1, 1, 1
C. __Al(s) + __H2SO4(aq)
+ __CO2(g)
__Al2O3(s)
3) 3, 3, 3, 1
__Al2(SO4)3(aq) + __H2(g)
2) 2, 3, 1, 3
3) 2, 3, 2, 3
17
Lecture 4.0: Chemical Reactions
and Quantities
Types of Reactions
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18
Type of Reactions
Chemical reactions can be classified as
 Combination reactions.
 Decomposition reactions.
 Single Replacement reactions.
 Double Replacement reactions.
19
Combination
In a combination reaction,
 Two or more elements (or simple compounds)
combine to form one product
A
+
B
A
B
2Mg(s) + O2(g)
2Na(s) + Cl2(g)
2MgO(s)
2NaCl(s)
SO3(g) + H2O(l)
H2SO4(aq)
20
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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21
Single Replacement
In a single replacement reaction,
 One element takes the place of a different element in
a reacting compound.
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Zn(s) + 2HCl(aq)
ZnCl2(aq) + H2(g)
Fe(s) + CuSO4(aq)
FeSO4(aq) + Cu(s)
22
Zn and HCl is a Single Replacement
Reaction
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23
Double Replacement
In a double replacement,
 Two elements in the reactants exchange places.
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AgNO3(aq) + NaCl(aq)
AgCl(s) + NaNO3(aq)
ZnS(s)
ZnCl2(aq) + H2S(g)
+ 2HCl(aq)
24
Example of a Double
Replacement
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25
Chemical Reactions and
Quantities
Oxidation-Reduction Reactions
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26
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)
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27
Electron Loss and Gain
An oxidation-reduction reaction
 Transfers electrons from one reactant to another.
A Loss of Electrons is Oxidation
Zn(s)
Zn2+(aq) + 2e-
(LEO)
A Gain of Electrons is Reduction
Cu2+(aq) + 2eCu(s)
(GER)
28
Oxidation and Reduction
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29
Zn and Cu2+
oxidation
Zn(s)
Silvery metal
Zn2+(aq) + 2ereduction
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Cu2+(aq) + 2eBlue
Cu(s)
orange
30
Electron Transfer from Zn to
Cu2+
Oxidation: electron loss
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Reduction: electron gain
31
Writing Oxidation and
Reduction Reactions
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) + 2e2F−(s)
reduction
32
Pause for ALE exercise
Do, in groups, problems from the active learning
exercise sheet
33
Lecture 4.0: Chemical Reactions
and Quantities
The Mole
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34
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
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35
A Mole of Atoms
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
36
A Mole of a Compound
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
37
Particle in One-Mole Samples
TABLE 5.3
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38
Avogadro’s Number
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
1 mole
and
1 mole
6.02 x 1023 particles
39
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
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40
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
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
42
Subscripts State Atoms and
Moles
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1 mole C9H8O4
=
9 moles C 8 moles H
4 moles O
43
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
moles O
1 mole C9H8O4
mole C9H8O4
8 moles H
4
1 mole C9H8O4
1
and
1 mole C9H8O4
1 mole C9H8O4
44
1 mole C9H8O4
Lecture 4.0: Chemical Reactions and
Quantities
Molar Mass
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45
Molar Mass
The molar mass is
 The mass of one mole of a substance.
 The atomic mass of an element expressed in grams.
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46
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
47
Molar Mass of K3PO4
Determine the molar mass of K3PO4 to 0.1 g.
Element
Number
of Moles
Atomic Mass
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
Total Mass in
K3PO4
212.3 g
48
One-Mole Quantities
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32.1 g
342.3 g
55.9 g
58.5 g
294.2 g
49
Conversion Factors from Molar
Mass
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
50
Calculations Using Molar Mass
 Mole factors are used to convert between the grams of a
substance and the number of moles.
Grams
Mole factor
Moles
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51
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
27.0 g Al
1 mole Al
= 81.0 g Al
mole factor for Al
52
Lecture 4.0 Chemical Reactions
and Quantities
Mole Relationships in
Chemical Equations
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53
Law of Conservation of Mass
The Law of Conservation of Mass indicates that
in an
ordinary chemical reaction,
 Matter cannot be created or destroyed.
 No change in total mass occurs in a reaction.
 Mass of products is equal to mass of
reactants.
54
Conservation of Mass
+
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Reactants
2 moles Ag
+ 1 moles S
2 (107.9 g)
+ 1(32.1 g)
247.9
=
=
Products
1 mole Ag2S
1 (247.9 g)
=
247.9 g
55
Reading Equations with Moles
Consider the following equation:
4Fe(s)
+ 3O2(g)
2Fe2O3(s)
This equation can be read in “moles” by placing the
word “moles” between each coefficient and formula.
4 moles Fe +
3 moles O2
2 moles Fe2O3
56
Writing Mole-Mole Factors
A mole-mole factor is a ratio of the moles for any two
substances in an equation.
4Fe(s)
+
Fe and O2
Fe and Fe2O3
O2 and Fe2O3
3O2(g)
2Fe2O3(s)
4 moles Fe
and
3 moles O2
4 moles Fe
and
2 moles Fe2O3
3 moles O2
and
2 moles Fe2O3
3 moles O2
4 moles Fe
2 moles Fe2O3
4 moles Fe
2 moles Fe2O3
3 moles O2
57
Calculations with Mole Factors
How many moles of Fe2O3 can be produced from
6.0 moles 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 moles Fe2O3
3 mole O2
58
Lecture 4.0: Chemical Reactions
and Quantities
Mass Calculations for Reactions
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59
Moles to Grams
Suppose we want to determine the mass (g) of NH3
that can form from 2.50 moles N2.
N2(g) + 3H2(g)
2NH3(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
60
Moles to Grams
The setup for the solution would be:
2.50 mole N2 x 2 moles NH3 x 17.0 g NH3
1 mole N2
1 mole NH3
given
mole-mole factor
molar mass
= 85.0 g NH3
61
Calculating the Mass of a
Reactant
The reaction between H2 and O2 produces 13.1 g water.
How many grams of O2 reacted?
2H2(g)
+ O2(g)
2H2O(g)
?g
13.1 g
The plan and factors would be
g H2O
mole H2O
mole O2
g O2
molar
mole-mole
molar
mass H2O
factor
mass O2
62
Calculating the Mass of a
Reactant
The setup would be:
13.1 g H2O x 1 mole H2O x 1 mole O2 x 32.0 g O2
18.0 g H2O
2 moles H2O 1 mole O2
molar
mole-mole
mass H2O
factor
molar
mass O2
= 11.6 g O2
63
Calculating the Mass of Product
When 18.6 g ethane gas C2H6 burns in oxygen, how
many grams of CO2 are produced?
2C2H6(g) + 7O2(g)
4CO2(g) + 6H2O(g)
18.6 g
?g
The plan and factors would be
g C2H6
mole C2H6
molar
mass C2H6
mole CO2
mole-mole
factor
g CO2
molar
mass CO2
64
Calculating the Mass of Product
The setup would be
18.6 g C2H6 x 1 mole C2H6 x 4 moles CO2 x 44.0 g CO2
30.1 g C2H6
2 moles C2H6 1 mole CO2
molar
mass C2H6
=
mole-mole
factor
molar
mass CO2
54.4 g CO2
65
Active learning exercise
66
Lecture 4.0: Chemical Reactions
and Quantities
Percent Yield and Limiting
Reactants
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67
Theoretical, Actual, and Percent
Yield
Theoretical yield
 The maximum amount of product, which is calculated
using the balanced equation.
Actual yield
 The amount of product obtained when the reaction
takes place.
Percent yield
 The ratio of actual yield to theoretical yield.
percent yield =
actual yield (g)
x 100
theoretical yield (g)
68
Calculating 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 burn and you throw them out. The rest
of the cookies are okay. What is the percent yield of
edible cookies?
Theoretical yield 60 cookies possible
Actual yield
48 cookies to eat
Percent yield
48 cookies x 100 = 80% yield
60 cookies
69
Limiting Reactant
A limiting reactant in a chemical reaction
is the
substance that
 Is used up first.
 Limits the amount of product that can
form and stops the reaction.
70
Reacting Amounts
In a table setting, there is 1plate,
1 fork, 1 knife, and 1 spoon.
How many table settings are
possible from 5 plates, 6 forks,
4 spoons, and 7 knives?
What is the limiting item?
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71
Reacting Amounts
Only 4 place settings are possible.
Initially Used Left over
Plates
5
4
1
Forks
6
4
2
Spoons 4
4
0
Knives 7
4
3
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The limiting item is the spoon.
72
Example of An Everyday
Limiting Reactant
How many peanut butter sandwiches could
be made from 8 slices of bread and 1 jar of
peanut butter?
With 8 slices of bread, only 4 sandwiches
could be made. The bread is the limiting
item.
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73
Example of An Everyday
Limiting Reactant
How many peanut butter sandwiches could
be made from 8 slices bread and 1
tablespoon of peanut butter?
With 1 tablespoon of peanut butter, only 1
sandwich could be made. The peanut butter
is the limiting item.
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74
Limiting Reactant
When 4.00 moles H2 is mixed with 2.00 moles
Cl2,how many moles of HCl can form?
H2(g) + Cl(g)  2HCl (g)
4.00 moles 2.00 moles
??? moles
 Calculate the moles of product that each
reactant, H2 and Cl2, could produce.
 The limiting reactant is the one that produces the
smallest amount of product.
75
Limiting Reactant
HCl from H2
4.00 moles H2 x 2 moles HCl = 8.00 moles HCl
1 moles H2
(not possible)
HCl from Cl2
2.00 moles Cl2 x 2 moles HCl = 4.00 moles HCl
1 mole Cl2 (smaller number of moles,
Cl2 will be used up first)
The limiting reactant is Cl2 because it is used up first.
Thus Cl2 produces the smaller number of moles of HCl.
76
Check Calculations
Initially
Reacted/
Formed
Left after
reaction
H2
Cl2
2HCl
4.00
moles
-2.00
moles
2.00
moles
-2.00
moles
2.00
moles
Excess
0
mole 4.00
moles
Limiting
0 mole
+4.00
moles
77
Limiting Reactants Using Mass
If 4.80 moles Ca are mixed with 2.00 moles N2, which is
The limiting reactant? 3Ca(s) + N2(g)  Ca3N2(s)
moles of Ca3N2 from Ca
4.80 moles Ca x 1 mole Ca3N2 = 1.60 moles Ca3N2
3 moles Ca
(Ca is used up)
moles of Ca3N2 from N2
2.00 moles N2 x 1 mole Ca3N2 = 2.00 moles Ca3N2
1 mole N2
(not possible)
Ca is used up when 1.60 mole Ca3N2 forms. Thus, Ca is
the limiting reactant.
78
Limiting Reactants Using Mass
Calculate the mass of water produced when 8.00
g H2
and 24.0 g O2 react?
2H2(g) + O2(g)
2H2O(l)
79
Limiting Reactants Using Mass
Moles H2O from H2:
8.00 g H2 x 1 mole H2 x 2 moles H2O = 3.97 moles H2O
2.02 g H2
2 moles H2
(not possible)
Moles H2O from O2:
24.0 g O2 x 1 mole O2 x 2 moles H2O = 1.50 moles H2O
32.0 g O2
1 mole O2
O2 is limiting
The maximum amount of product is 1.50 moles H2O,
which is converted to grams.
1.50 moles H2O x 18.0 g H2O = 27.0 g H2O
1 mole H2O
80
More Active Learning Exercises
81