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8/4/2019 Chapter 08au
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Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Introductory Chemistry, 2nd Edition
Nivaldo Tro
Chapter 8
Quantities inChemical Reactions
2006, Prentice Hall
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Tro's Introductory Chemistry, Chapter8
2
Quantities in Chemical Reactions
the amount of every substance used andmade in a chemical reaction is related to theamounts of all the other substances in the
reactionLaw of Conservation of Mass
balancing equations by balancing atoms
the study of the numerical relationshipbetween chemical quantities in a chemicalreaction is called reactionstoichiometry
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Tro's Introductory Chemistry, Chapter8
3
Global Warming
scientists have measured an average 0.6Crise in atmospheric temperature since 1860
during the same period atmospheric CO2
levels have risen 25% are the two trends causal?
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Tro's Introductory Chemistry, Chapter8
4
The Source of Increased CO2
the primary source of the increased CO2levels are combustion reactions of fossilfuels we use to get energy
1860 corresponds to the beginning of the
Industrial Revolution in the US and Europe
(g)(g)(g)(g) OH2COO2CH 2224 (g)(g)(g)(l) OH18CO16O25HC2 222188
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Tro's Introductory Chemistry, Chapter8
6
Making Pancakes
if you want to make more or less than 5
pancakes you can use the number of eggs you
have to determine the number of pancakes you
can makeassuming you have enough flour and baking powder
pancakes20eggs2
pancakes5eggs8
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Tro's Introductory Chemistry, Chapter8
7
Making Molecules
Mole-to-Mole Conversions the balanced equation is the recipe for a chemical
reaction
the equation 3 H2(g) + N2(g)
2 NH3(g) tells us that3 molecules of H2 react with exactly 1 molecule of N2and make exactly 2 molecules of NH3 or3 molecules H2 1 molecule N2 2 molecules NH3
in this reaction
and since we count molecules by moles
3 moles H2 1 mole N2 2 moles NH3
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Example 8.1
Mole-to-Mole Conversions
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Tro's Introductory Chemistry, Chapter8
9
Example:
Sodium chloride, NaCl, forms by the following reactionbetween sodium and chlorine. How many moles of NaCl
result from the complete reaction of 3.4 mol of Cl2?
Assume there is more than enough Na.
2 Na(s) + Cl2(g) 2 NaCl(s)
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Tro's Introductory Chemistry, Chapter8
10
Example:
How many moles of NaCl
result from the complete
reaction of 3.4 mol of Cl2
in
the reaction below?
2 Na(s) + Cl2(g) 2 NaCl(s)
Write down the given quantity and its units.
Given: 3.4 mol Cl2
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Tro's Introductory Chemistry, Chapter8
11
Write down the quantity to find and/or its units.
Find: ? moles NaCl
Information
Given: 3.4 mol Cl2
Example:
How many moles of NaCl
result from the complete
reaction of 3.4 mol of Cl2
in
the reaction below?
2 Na(s) + Cl2(g) 2 NaCl(s)
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Tro's Introductory Chemistry, Chapter8
12
Collect Needed Conversion Factors:
according to the equation:
1 mole Cl2
2 moles NaCl
Information
Given: 3.4 mol Cl2
Find: ? moles NaCl
Example:
How many moles of NaCl
result from the complete
reaction of 3.4 mol of Cl2
in
the reaction below?
2 Na(s) + Cl2(g) 2 NaCl(s)
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Tro's Introductory Chemistry, Chapter8
13
Write a Solution Map for converting the units :
mol Cl2 mol NaCl
2Clmol1
NaClmol2
Information
Given: 3.4 mol Cl2
Find: ? moles NaCl
CF: 1 mol Cl2 2 mol NaCl
Example:
How many moles of NaCl
result from the complete
reaction of 3.4 mol of Cl2
in
the reaction below?
2 Na(s) + Cl2(g) 2 NaCl(s)
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Tro's Introductory Chemistry, Chapter8
14
Apply the Solution Map:
NaClmolesClmol1
NaClmol2Clmol43
22 .
= 6.8 mol NaCl
= 6.8 moles NaCl Sig. Figs. & Round:
Information
Given: 3.4 mol Cl2
Find: ? moles NaCl
CF: 1 mol Cl2 2 mol NaCl
SM: mol Cl2 mol NaCl
Example:
How many moles of NaCl
result from the complete
reaction of 3.4 mol of Cl2
in
the reaction below?
2 Na(s) + Cl2(g) 2 NaCl(s)
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Tro's Introductory Chemistry, Chapter8
15
Check the Solution:
3.4 mol Cl2 6.8 mol NaCl
The units of the answer, moles NaCl, are correct.
The magnitude of the answer makes sense
since the equation tells us you make twice as manymoles of NaCl as the moles of Cl2 .
Information
Given: 3.4 mol Cl2
Find: ? moles NaCl
CF: 1 mol Cl2 2 mol NaCl
SM: mol Cl2 mol NaCl
Example:
How many moles of NaCl
result from the complete
reaction of 3.4 mol of Cl2
in
the reaction below?
2 Na(s) + Cl2(g) 2 NaCl(s)
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16
Making Molecules
Mass-to-Mass Conversions we know there is a relationship between the mass and numberof moles of a chemical
1 mole = Molar Mass in grams
the molar mass of the chemicals in the reaction and thebalanced chemical equation allow us to convert from the
amount of any chemical in the reaction to the amount of any
other
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Example 8.2
Mass-to-Mass Conversions
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Tro's Introductory Chemistry, Chapter8
18
Example:
In photosynthesis, plants convert carbon dioxide andwater into glucose, (C6H12O6), according to the following
reaction. How many grams of glucose can be synthesized
from 58.5 g of CO2? Assume there is more than enough
water to react with all the CO2.
(aq)(g)(l)(g) 61262sunlight
22 OHCO6OH6CO6
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Tro's Introductory Chemistry, Chapter8
19
Example:
How many grams of glucose
can be synthesized from 58.5 g
of CO2
in the reaction?
6 CO2(g) + 6 H2O(l)
6 O2(g) + C6H12O6(aq)
Write down the given quantity and its units.
Given: 58.5 g CO2
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Tro's Introductory Chemistry, Chapter8
20
Write down the quantity to find and/or its units.
Find: ? g C6H12O6
Information
Given: 55.4 g CO2
Example:
How many grams of glucose
can be synthesized from 58.5 g
of CO2
in the reaction?
6 CO2(g) + 6 H2O(l)
6 O2(g) + C6H12O6(aq)
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21
Collect Needed Conversion Factors:
Molar Mass C6H12O6 = 6(mass C) + 12(mass H) + 6(mass O)
= 6(12.01) + 12(1.01) + 6(16.00) = 180.2 g/mol
Molar Mass CO2 = 1(mass C) + 2(mass O)
= 1(12.01) + 2(16.00) = 44.01 g/mol1 mole CO2 = 44.01 g CO2
1 mole C6H12O6 = 180.2 g C6H12O6
1 mole C6H12O6 6 mol CO2 (from the chem. equation)
Information
Given: 55.4 g CO2
Find: g C6H12O6
Example:
How many grams of glucose
can be synthesized from 58.5 g
of CO2
in the reaction?
6 CO2(g) + 6 H2O(l)
6 O2(g) + C6H12O6(aq)
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Tro's Introductory Chemistry, Chapter8
22
Write a Solution Map:
g
CO2
6126
6126
OHCmol1
OHCg80.21
2
2
COg4.014
COmol1
Information
Given: 58.5 g CO2
Find: g C6H12O6
CF: 1 mol C6H12O6 = 180.2 g1 mol CO2 = 44.01 g
1 mol C6H12O6 6 mol CO2
mol
CO2
mol
C6H12O6
g
C6H12O6
2
6126
COmol6
OHCmol1
Example:
How many grams of glucose
can be synthesized from 58.5 g
of CO2
in the reaction?
6 CO2(g) + 6 H2O(l)
6 O2(g) + C6H12O6(aq)
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Tro's Introductory Chemistry, Chapter8
23
6126
6126
2
6126
2
22
OHCmole1
OHCg80.21
COmol6
OHCmol1
COg44.01
COmole1COg8.55
Apply the Solution Map:
= 39.9216 g C6H12O6
= 39.9 g C6H12O6
Sig. Figs. & Round:
Information
Given: 58.5 g CO2
Find: g C6H12O6
CF: 1 mol C6H12O6 = 180.2 g1 mol CO2 = 44.01 g
1 mol C6H12O6 6 mol CO2
SM: g CO2 mol CO2mol C6H12O6 g C6H12O6
Example:
How many grams of glucose
can be synthesized from 58.5 g
of CO2
in the reaction?
6 CO2(g) + 6 H2O(l)
6 O2(g) + C6H12O6(aq)
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Tro's Introductory Chemistry, Chapter8
24
Check the Solution:
58.5 g CO2 = 39.9 g C6H12O6
The units of the answer, g C6H12O6, are correct.It is hard to judge the magnitude.
Information
Given: 58.5 g CO2
Find: g C6H12O6
CF: 1 mol C6H12O6 = 180.2 g1 mol CO2 = 44.01 g
1 mol C6H12O6 6 mol CO2
SM: g CO2 mol CO2mol C6H12O6 g C6H12O6
Example:
How many grams of glucose
can be synthesized from 58.5 g
of CO2 in the reaction?
6 CO2(g) + 6 H2O(l)
6 O2(g) + C6H12O6(aq)
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Tro's Introductory Chemistry, Chapter8
25
More Making Pancakes we know that
but what would happen if we had 3 cups of flour,
10 eggs, and 4 tsp of baking powder?
1 cup Flour + 2 Eggs + tsp Baking Powder 5 Pancakes
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Tro's Introductory Chemistry, Chapter8
26
More Making Pancakes
pancakes25eggs2
pancakes5eggs10
pancakes15flourcu1
pancakes5flourcu3
pancakes40powbaktsp
21pancakes5powbaktsp4
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Tro's Introductory Chemistry, Chapter8
27
More Making Pancakes each ingredient could potentially make a different
number of pancakes
but all the ingredients have to work together!
we only have enough flour to make 15 pancakes, so
once we make 15 pancakes the flour runs out, nomatter how much of the other ingredients we have
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Tro's Introductory Chemistry, Chapter8
28
More Making Pancakes
The flour limits the amount of pancakes we can
make. In chemical reactions we call this the
limiting reactant.
also known as limiting reagent
The maximum number of pancakes we can make
depends on this ingredient. In chemical reactions
we call this the theoretical yield.
it also determines the amounts of the other ingredients
we will use!
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29
More Making Pancakes Lets now assume that as we are making pancakes we
spill some of the batter, burn a pancake, drop one onthe floor or other uncontrollable events happen so thatwe only make 11 pancakes. The actual amount ofproduct made in a chemical reaction is called the
actual yield. We can determine the efficiency of our pancake-
making by calculating the percentage of themaximum number of pancakes we actually make. In
chemical reactions we call this the percent yield.
YieldPercent100YieldlTheoretica
YieldActual % %% 37100
pancakes15
pancakes11
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Tro's Introductory Chemistry, Chapter8 30
Theoretical and Actual Yield
As we did with the pancakes, in order to determinethe theoretical yield, we should use reactionstoichiometry to determine the amount of producteach of our reactants could make.
The theoretical yield will always be the leastpossible amount of product.
The theoretical yield will always come from thelimiting reactant.
Because of both controllable and uncontrollablefactors, the actual yield of product will always beless than the theoretical yield.
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Example 8.6
Finding Limiting Reactant,
Theoretical Yield andPercent Yield
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Tro's Introductory Chemistry, Chapter8 32
Example:
When 11.5 g of C are allowed to react with 114.5 g ofCu2O in the reaction below, 87.4 g of Cu are obtained.
Find the Limiting Reactant, Theoretical Yield and
Percent Yield.
(g)(s)(s)(s) COCu2CuOCu 2
E l
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Tro's Introductory Chemistry, Chapter8 33
Example:
When 11.5 g of C reacts with
114.5 g of Cu2O, 87.4 g of Cu are
obtained. Find the Limiting
Reactant, Theoretical Yield and
Percent Yield.
Cu2O(s) + C(s) 2 Cu(s) + CO(g)
Write down the given quantity and its units.
Given: 11.5 g C
114.5 g Cu2O
87.4 g Cu produced
I f tiE l
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Tro's Introductory Chemistry, Chapter8 34
Write down the quantity to find and/or its units.
Find: limiting reactant
theoretical yield
percent yield
Information
Given: 11.5 g C, 114.5 g Cu2O
87.4 g Cu produced
Example:
When 11.5 g of C reacts with
114.5 g of Cu2O, 87.4 g of Cu are
obtained. Find the Limiting
Reactant, Theoretical Yield and
Percent Yield.
Cu2O(s) + C(s) 2 Cu(s) + CO(g)
InformationE l
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35
Collect Needed Conversion Factors:Molar Mass Cu2O = 143.02 g/mol
Molar Mass Cu = 63.54 g/mol
Molar Mass C = 12.01 g/mol
1 mole Cu2O 2 mol Cu (from the chem. equation)
1 mole C 2 mol Cu (from the chem. equation)
Information
Given: 11.5 g C, 114.5 g Cu2O
87.4 g Cu produced
Find: Lim. Rct., Theor. Yld., % Yld.
Example:
When 11.5 g of C reacts with
114.5 g of Cu2O, 87.4 g of Cu are
obtained. Find the Limiting
Reactant, Theoretical Yield and
Percent Yield.
Cu2O(s) + C(s) 2 Cu(s) + CO(g)
InformationE l
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Tro's Introductory Chemistry, Chapter8 36
Write a Solution Map:
Information
Given: 11.5 g C, 114.5 g Cu2O
87.4 g Cu produced
Find: Lim. Rct., Theor. Yld., % Yld.
CF: 1 mol C = 12.01 g; 1 mol Cu =63.54 g; 1 mol Cu2O = 143.08 g;1 mol Cu2O 2 mol Cu; 1 mol C 2 mol Cu
Example:
When 11.5 g of C reacts with
114.5 g of Cu2O, 87.4 g of Cu are
obtained. Find the Limiting
Reactant, Theoretical Yield and
Percent Yield.
Cu2O(s) + C(s) 2 Cu(s) + CO(g)
g
Cu2O
g
C
Cumol1
Cug3.546
Cumol1
Cug3.546
OCug.08143
OCumol1
2
2
Cg.0112
Cmol1
mol
Cu2O
mol
C
mol
Cu
mol
Cu
g
Cu
g
Cu
OCumol1
Cumol2
2
Cmol1
Cumol2 }smallest
amount is
theoretical
yield
InformationExample:
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Tro's Introductory Chemistry, Chapter8 37
Cug7101Cumole1Cug3.546
OCumol1Cumol2
OCug143.08OCumole1OCug14.51
22
22 .
Apply the Solution Map:
Information
Given: 11.5 g C, 114.5 g Cu2O
87.4 g Cu produced
Find: Lim. Rct., Theor. Yld., % Yld.
CF: 1 mol C = 12.01 g; 1 mol Cu =63.54 g; 1 mol Cu2O = 143.08 g;1 mol Cu2O 2 mol Cu; 1 mol C 2 mol Cu
SM: g rct mol rct mol Cu g Cu
Example:
When 11.5 g of C reacts with
114.5 g of Cu2O, 87.4 g of Cu are
obtained. Find the Limiting
Reactant, Theoretical Yield and
Percent Yield.
Cu2O(s) + C(s) 2 Cu(s) + CO(g)
Cug122Cumole1
Cug3.546
Cmol1
Cumol2
Cg12.01
Cmole1Cg1.51
InformationExample:
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Tro's Introductory Chemistry, Chapter8 38
Apply the Solution Map:
114.5 g Cu2O can make 101.7 g Cu
11.5 g C can make 122 g Cu Theoretical Yield = 101.7 g Cu
Limiting Reactant = Cu2O
Information
Given: 11.5 g C, 114.5 g Cu2O
87.4 g Cu produced
Find: Lim. Rct., Theor. Yld., % Yld.
CF: 1 mol C = 12.01 g; 1 mol Cu =63.54 g; 1 mol Cu2O = 143.08 g;1 mol Cu2O 2 mol Cu; 1 mol C 2 mol Cu
SM: g rct mol rct mol Cu g Cu
Example:
When 11.5 g of C reacts with
114.5 g of Cu2O, 87.4 g of Cu are
obtained. Find the Limiting
Reactant, Theoretical Yield and
Percent Yield.
Cu2O(s) + C(s) 2 Cu(s) + CO(g)
Least Amount
114.5 g Cu2O can make101.7 g Cu
InformationExample:
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Tro's Introductory Chemistry, Chapter8 39
Write a Solution Map:
Information
Given: 11.5 g C, 114.5 g Cu2O
87.4 g Cu produced
Find: Lim. Rct., Theor. Yld., % Yld.
CF: 1 mol C = 12.01 g; 1 mol Cu =63.54 g; 1 mol Cu2O = 143.08 g;1 mol Cu2O 2 mol Cu; 1 mol C 2 mol Cu
Example:
When 11.5 g of C reacts with
114.5 g of Cu2O, 87.4 g of Cu are
obtained. Find the Limiting
Reactant, Theoretical Yield and
Percent Yield.
Cu2O(s) + C(s) 2 Cu(s) + CO(g)
YieldPercent100
YieldlTheoretica
YieldActual %
InformationExample:
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Tro's Introductory Chemistry, Chapter8 40
Apply the Solution Map:
Information
Given: 11.5 g C, 114.5 g Cu2O
87.4 g Cu produced
Find: Lim. Rct., Theor. Yld., % Yld.
CF: 1 mol C = 12.01 g; 1 mol Cu =63.54 g; 1 mol Cu2O = 143.08 g;1 mol Cu2O 2 mol Cu; 1 mol C 2 mol Cu
SM:
Example:
When 11.5 g of C reacts with
114.5 g of Cu2O, 87.4 g of Cu are
obtained. Find the Limiting
Reactant, Theoretical Yield and
Percent Yield.
Cu2O(s) + C(s) 2 Cu(s) + CO(g)
YieldPercent100YieldlTheoretica
YieldActual %
%.% 958100Cug101.7Cug87.4
YieldPercent100YieldlTheoretica
YieldActual %
InformationExample:
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T ' I d Ch i Ch 41
Check the Solutions:
Limiting Reactant = Cu2O
Theoretical Yield = 101.7 g
Percent Yield = 85.9%
The Percent Yield makes sense as it is less than 100%.
Information
Given: 11.5 g C, 114.5 g Cu2O
87.4 g Cu produced
Find: Lim. Rct., Theor. Yld., % Yld.
CF: 1 mol C = 12.01 g; 1 mol Cu =63.54 g; 1 mol Cu2O = 143.08 g;1 mol Cu2O 2 mol Cu; 1 mol C 2 mol Cu
Example:
When 11.5 g of C reacts with
114.5 g of Cu2O, 87.4 g of Cu are
obtained. Find the Limiting
Reactant, Theoretical Yield and
Percent Yield.
Cu2O(s) + C(s) 2 Cu(s) + CO(g)