Chapter 08au

8/4/2019 Chapter 08au

1/41

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
http://encarta.msn.com/encnet/features/dictionary/DictionaryResults.aspx?search=stoichiometryhttp://encarta.msn.com/encnet/features/dictionary/DictionaryResults.aspx?search=stoichiometry


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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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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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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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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
http://wps.prenhall.com/wps/media/objects/1053/1078985/media/AACXJQT0.htmlhttp://wps.prenhall.com/wps/media/objects/1053/1078985/media/AACXJQT0.htmlhttp://wps.prenhall.com/wps/media/objects/1053/1078985/media/AACXJQT0.htmlhttp://wps.prenhall.com/wps/media/objects/1053/1078985/media/AACXJQT0.htmlhttp://wps.prenhall.com/wps/media/objects/1053/1078985/media/AACXJQT0.htmlhttp://wps.prenhall.com/wps/media/objects/1053/1078985/media/AACXJQT0.html


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Example 8.1

Mole-to-Mole Conversions


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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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10

Example:

How many moles of NaCl

result from the complete

reaction of 3.4 mol of Cl2

in

the reaction below?


Write down the given quantity and its units.

Given: 3.4 mol Cl2


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11

Write down the quantity to find and/or its units.

Find: ? moles NaCl

Information

Given: 3.4 mol Cl2

Example:




in

the reaction below?



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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:




in

the reaction below?



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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:




in

the reaction below?



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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


SM: mol Cl2 mol NaCl

Example:




in

the reaction below?



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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


SM: mol Cl2 mol NaCl

Example:




in

the reaction below?



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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
http://wps.prenhall.com/wps/media/objects/165/169519/StoichiometryCalculation.htmlhttp://wps.prenhall.com/wps/media/objects/165/169519/StoichiometryCalculation.htmlhttp://wps.prenhall.com/wps/media/objects/165/169519/StoichiometryCalculation.htmlhttp://wps.prenhall.com/wps/media/objects/165/169519/StoichiometryCalculation.htmlhttp://wps.prenhall.com/wps/media/objects/165/169519/StoichiometryCalculation.htmlhttp://wps.prenhall.com/wps/media/objects/165/169519/StoichiometryCalculation.html


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Example 8.2

Mass-to-Mass Conversions


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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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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)


Given: 58.5 g CO2


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20


Find: ? g C6H12O6

Information

Given: 55.4 g CO2

Example:



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:



of CO2

in the reaction?

6 CO2(g) + 6 H2O(l)

6 O2(g) + C6H12O6(aq)


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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:



of CO2

in the reaction?

6 CO2(g) + 6 H2O(l)

6 O2(g) + C6H12O6(aq)


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23

6126

6126

2

6126

2

22

OHCmole1

OHCg80.21

COmol6

OHCmol1

COg44.01

COmole1COg8.55


= 39.9216 g C6H12O6

= 39.9 g C6H12O6

Sig. Figs. & Round:

Information

Given: 58.5 g CO2

Find: g C6H12O6



SM: g CO2 mol CO2mol C6H12O6 g C6H12O6

Example:



of CO2

in the reaction?

6 CO2(g) + 6 H2O(l)

6 O2(g) + C6H12O6(aq)


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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



SM: g CO2 mol CO2mol C6H12O6 g C6H12O6

Example:



of CO2 in the reaction?

6 CO2(g) + 6 H2O(l)

6 O2(g) + C6H12O6(aq)


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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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26

More Making Pancakes

pancakes25eggs2

pancakes5eggs10

pancakes15flourcu1

pancakes5flourcu3

pancakes40powbaktsp

21pancakes5powbaktsp4


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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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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!
http://wps.prenhall.com/wps/media/objects/165/169519/LimitingReactant.htmlhttp://wps.prenhall.com/wps/media/objects/165/169519/LimitingReactant.html


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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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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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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)


Given: 11.5 g C

114.5 g Cu2O

87.4 g Cu produced

I f tiE l


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Find: limiting reactant

theoretical yield

percent yield

Information

Given: 11.5 g C, 114.5 g Cu2O

87.4 g Cu produced

Example:





Percent Yield.


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:





Percent Yield.


InformationE l


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Information

Given: 11.5 g C, 114.5 g Cu2O

87.4 g Cu produced


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:





Percent Yield.


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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Cug7101Cumole1Cug3.546

OCumol1Cumol2

OCug143.08OCumole1OCug14.51

22

22 .


Information

Given: 11.5 g C, 114.5 g Cu2O

87.4 g Cu produced



SM: g rct mol rct mol Cu g Cu

Example:





Percent Yield.


Cug122Cumole1

Cug3.546

Cmol1

Cumol2

Cg12.01

Cmole1Cg1.51

InformationExample:


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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



SM: g rct mol rct mol Cu g Cu

Example:





Percent Yield.


Least Amount

114.5 g Cu2O can make101.7 g Cu

InformationExample:


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Information

Given: 11.5 g C, 114.5 g Cu2O

87.4 g Cu produced



Example:





Percent Yield.


YieldPercent100

YieldlTheoretica

YieldActual %

InformationExample:


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Information

Given: 11.5 g C, 114.5 g Cu2O

87.4 g Cu produced



SM:

Example:





Percent Yield.



YieldActual %

%.% 958100Cug101.7Cug87.4


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



Example:





Percent Yield.


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