5 - 1CH104 CHAPTER 7 Chemical Reactions & Quantities Reactions & Equations Balancing Chemical Reactions Types of Reactions Oxidation-Reduction The Mole.

5 - 1CH104

CHAPTER 7Chemical Reactions & Quantities

Reactions & Equations

Balancing Chemical Reactions

Types of Reactions

Oxidation-Reduction

The Mole and Chemical Equations

Mass Calculations

Percent Yield & Limiting Reactants

Energy Changes in Reactions

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Examples

Color

Odor

Taste

Feel

Shape

Physical propertiesCharacteristics that can be evaluated

without changing the composition of the material.

Density

Melting / Freezing point

Boiling point

Compressibility

Form (foil, wire, powder…)

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What is Chemistry?“The study of Matter and its Changes.”

Physical Changes =

Changes in a Physical Property

Chemical Changes =

Changes in a Chemical Property

Appearance: • melting, freezing, evaporation…• stretching, molding, cutting…

Chemical Composition:

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Change in the Chemical Composition

Burning of Magnesium

Chemical Changes

Rusting of Iron

Decomposing of wood

Souring of Milk

Examples:

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Examples

Mulching leaves

Which are chemical or physical changes?Which are chemical or physical changes?

Tarnishing Silver

Fermentation

Making ice into water

Carbonated Beverage going flat

Bleaching a stain

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Mg + O2 MgO +

Energy

Chemical Reactions

Shows how the Chemical change occurs.Shows how the Chemical change occurs.

Reactants

C3H8 + O2 CO2 + H2O + Energy

Fe + O2 Fe2O3

ProductsProducts

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Chemical equations

Chemist’s shorthand to describe a reaction.

•Reactants •Products

•The state of all substances

H2 + O2 H2O + E(g) (g) (g)

•Any conditions used in the reaction

heat

•Same # & type atoms on each side• Law of Conservation of Matter

2 2

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Balancing Equations

___W10 + ___B8 ___WB___W10 + ___B8 ___WB

•Reactants •Products

Making Hot dogs:How many packages wieners & buns to buy so none is left over.

4 5 40

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Ca

H

Cl

Balancing Equations

Ca + HCl CaCl2 + H2 Ca + HCl CaCl2 + H2

•Reactants •Products

Step 1: Count atoms of each element on both sides of equation.

1

1

1

1

2

2

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Balancing Equations

Ca + HCl CaCl2 + H2 Ca + HCl CaCl2 + H2

Ca

H

Cl

Ca

H

Cl

•Reactants •Products

1

1

1

1

1

1

1

2

2

1

2

2

Step 2: Determine which atoms are not balanced.

- not balanced

- not balanced

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Balancing Equations

Ca + HCl CaCl2 + H2 Ca + HCl CaCl2 + H2

•Reactants •Products

1

1

1

1

1

1

1

2

2

1

2

2

- not balanced

- not balanced

Step 3: Balance one element at a time with coefficients in front of formulas until all balanced.

(Never change the formula!)

2

22

Ca

H

Cl

Ca

H

Cl

22

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NaPO

MgCl

Balancing Equations

Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl

•Reactants •Products

Step 1: Count atoms of each element on both sides of equation.

31412

1 2831

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NaPO

MgCl

Balancing Equations

Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl

•Reactants •Products31412

1 2831

- not balanced

- not balanced

Step 2: Determine which atoms are not balanced.

- not balanced

- not balanced

- not balanced

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NaPO

MgCl

Balancing Equations

Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl

•Reactants •Products31412

1 2831

- not balanced

- not balanced

- not balanced

- not balanced

- not balanced

Step 3: Balance elements with #’s in front of formulas until all balanced.

(Never change the formulas!)

5 - 16CH104

Hints:• Start with a metal in a complex

compound, or an element that only appears in one formula. (Like Mg here)

NaPO

MgCl

Balancing Equations

Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl Na3PO4 + MgCl2 Mg3(PO4)2 + NaCl

•Reactants •Products31412

1 2831

- not balanced

- not balanced

- not balanced

- not balanced

- not balanced66

66

63

33

22

66

1

88

2

66

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Hints:• Start with an element that only appears

in one formula on both sides of the equation.

• Leave oxygen until last.

Balancing Equations

C2H6 + O2 CO2 + H2OC2H6 + O2 CO2 + H2O

•Reactants •ProductsC

H

O

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Balancing Equations

C2H6 + O2 CO2 + H2OC2H6 + O2 CO2 + H2O

C

H

O

C

H

O

•Reactants •Products

Step 1: Count atoms of each element on both sides of equation.

2

6

2

2

6

2

1

2

3

1

2

3

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C2H6 + O2 CO2 + H2OC2H6 + O2 CO2 + H2O

Balancing Equations

•Reactants •Products

2

6

2

2

6

2

1

2

3

1

2

3

Step 2: Determine which atoms are not balanced.

- not balanced

- not balanced

- not balancedC

H

O

C

H

O

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2

6

2

2

6

2

- not balanced

- not balanced

C2H6 + O2 CO2 + H2OC2H6 + O2 CO2 + H2O

22

Balancing Equations

•Reactants •Products

1

2

3

1

2

3

- not balanced

Step 3: Balance one element at a time with coefficients in front of formulas until all balanced.

(Never change the formula!)

2

55

3

66

77

3.5

77

C

H

O

C

H

O

5 - 21CH104

C

H

O

C

H

O

C2H6 + O2 CO2 + H2OC2H6 + O2 CO2 + H2O

22

Balancing Equations

•Reactants •Products

2

6

2

2

6

2

1

2

3

1

2

3

2

55

3

66

77

3.5

77

Can’t have 3.5 O2 , so multiply equation by 2!

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3.5C2H6 + O2 CO2 + H2OC2H6 + O2 CO2 + H2O

22

Balancing Equations

C

H

O

C

H

O

•Reactants •Products

2

6

2

2

6

2

1

2

3

1

2

3

4

55

6

66

77

7

77

Can’t have 3.5 O2 , so multiply equation by 2!

2

44 44

1212 1212

1414 1414

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(NH2)2CO + H2O ______> NH3 + CO2(NH2)2CO + H2O ______> NH3 + CO2

- not balanced

- not balanced

1

6

2 NHCO

NHCO

1

3

1 2 2

Balancing Equations

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(NH2)2CO + H2O ______> NH3 + CO2(NH2)2CO + H2O ______> NH3 + CO2

Balancing Equations

1

6

2 1

3

1

2

2

6NHCO

NHCO2 2

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Example: Decomposition of urea

CH3OH + PCl5 CH3Cl + POCl3 + H2O CH3OH + PCl5 CH3Cl + POCl3 + H2O

- not balanced

- not balanced

1

4

1 CHOPCl

CHOPCl

1

5

2 1 1

5 4

- not balanced

- not balanced

5 - 26CH104

1

CH3OH + PCl5 CH3Cl + POCl3 + H2O CH3OH + PCl5 CH3Cl + POCl3 + H2O

Balancing Equations

2

2

81

4

1 CHOPCl

CHOPCl

5

2 1 1

5 4 5

2

2

82

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Types of Chemical Reactions

Combination

Decomposition

Single Replacement: Substitution

Double Replacement: Metathesis

A + BX B + AX

A + B C

C A + B

AX + BY BX + AY

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Types of Chemical Reactions

Combination

Decomposition

Single Replacement: Substitution

Double Replacement: Metathesis

2H2 + O2 2H2O

CaCO3 CaO + CO2

Al + FeCl3 Fe + AlCl3

2AgNO3 + K2SO4 Ag2SO4 + 2KNO3

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Types of Chemical Reactions

Complete:C3H8 + 5O2 3CO2 + 4H2O

Combustion

Incomplete:2C3H8 + 7O2 6CO + 8H2O

C3H8 + 2O2 3C + 4H2O

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Combination Reactions

2H2 + O2 2H2O

Formation of Acid Rain

SO3 + H2O H2SO4

Explosion of Hydrogen Balloon

Rusting of Iron

4 Fe + 3 O2 2 Fe2O3

A + B C A + B C

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Decomposition Reactions

Heating Egg Shells

CaCO3 CaO + CO2

2 H2O2 2 H2O + O2

Blood with peroxide

C A + B

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Single Replacement Reactions

Iron Deposits on an Aluminum Pan

Al + FeCl3 Fe + AlCl3

A + BX B + AX

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Activity series of metalspotassiumsodium

potassiumsodium

calciumcalcium

magnesiumaluminum

zincchromium

magnesiumaluminum

zincchromium

ironnickel

tinlead

ironnickel

tinlead

coppersilver

platinumgold

coppersilver

platinumgold

Hydrogen

Al + Fe+3 Fe + Al+3

Fe + H+ Fe+3 + H2

incr

easi

ng

rea

ctiv

ity

Element give e’s to ion

lower on list

Element give e’s to ion

lower on list

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Double Replacement Reaction

BaCl2(aq) + Na2SO4(aq) BaSO4(s) + 2NaCl(aq)

AX + BY BX + AY AX + BY BX + AY

Ba+2 Cl-1

Na+1 SO4-2

Insoluble PrecipitateFormed

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Predict the products:

AgNO3(aq) + AlCl3 (aq)

Ag+ NO3

-

Al+3 Cl-

AgCl(s) + Al(NO3)3(aq)

Double Replacement ReactionAX + BY BX + AY AX + BY BX + AY

Balance later as needed to get:

3AgNO3(aq) + AlCl3 (aq) 3AgCl(s) + Al(NO3)3(aq)

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Oxidation and reductionREDOX

Where reactants exchange electrons -

Examples:

•All types of batteries

alkaline, NiCad, car batteries•Rusting and corrosion•Metabolism•Antioxidants (Vit C, E prevent oxidation)

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Oxidation and reductionREDOX

Where reactants exchange electrons -

Oxidation = Losing electronsLEO: Lose Electrons Oxidation

LEO the lion says GER

Reduction = Gaining electronsGER: Gain Electrons Reduction

OIL : Oxidation Is Losing

RIG : Reduction Is Gaining

OIL RIG

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Oxidation and reduction

2 Na(s) + Cl2 (g) 2 NaCl2 Na(s) + Cl2 (g) 2 NaCl

1-1+

Assign Oxidation States:

0 0For simple ions,

Ox state = charge.

For element in natural form

Ox State = 0.

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Oxidation and reduction

2 Na(s) + Cl2 (g) 2 NaCl2 Na(s) + Cl2 (g) 2 NaCl

1-1+

Who’s loosing or gaining electrons?

0 0

Loses 1 e- = LEO

Gains 1 e-=GER

Na loses e- (LEO) Na gets oxidized

Cl gains e- (GER) Cl gets reduced

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Oxidation and reduction

Oxidation - when reactant loses e-(s). (LEO)

Na (s) Na+ + e-

Reduction - when reactant gains e-(s). (GER)

Cl2 (g) + 2 e- 2 Cl-

These are half reactions

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2 Na(s) + Cl2 (g) + 2 e- 2 Na+ + 2e- + 2 Cl-

Oxidation and reduction

2 half reactions make a complete reaction

Na (s) Na+ + e-

Cl2 (g) + 2 e- 2 Cl-

2 2 2

2 Na(s) + Cl2 (g) 2 Na+ + 2 Cl-

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Oxidizing agent =• The chemical that caused an oxidation.• It is reduced.

Reducing agent =•The chemical that caused a reduction.• It is oxidized.

Oxidation and reduction

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Oxidation and reduction

2 Na(s) + Cl2 (g) 2 NaCl2 Na(s) + Cl2 (g) 2 NaCl

1-1+

Who’s loosing or gaining electrons?

0 0

• Na loses e- (LEO) so Na gets oxidized

• Na caused Cl to get reduced• Na is the Reducing agent

Loses 1 e- =LEO

Gains 1 e- =GER

5 - 44CH104

Oxidation and reduction

2 Na(s) + Cl2 (g) 2 NaCl2 Na(s) + Cl2 (g) 2 NaCl

1-1+

Who’s loosing or gaining electrons?

0 0

Loses 1 e- =LEO

Gains 1 e- =GER

• Cl gains e- (GER) so Cl gets reduced • Cl caused Na to get oxidized• Cl is the Oxidizing agent

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Oxidation state

Rules• Oxidation state of element in natural form = 0.

• For simple monoatomic ions, oxidation state = charge.

Describes the charge of each element.

• For certain groups at certain times, oxidation number = group number

Examples Na+1, Cl-1, Ca2+

Examples N+5, Cl+7

Examples N2, Na, O2, H2

5 - 47CH104

Hydrogen H1+ if bonded to nonmetal

H1- if bonded to metal

Example HCl

Oxidation statesOxygen O-2 Usually

O-1 in peroxides

Example H2O

Example H2O2

Example NaH

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Assign the oxidation states for all elements in

H2O

+1 -2

+1

+2 -2

Oxidation states

5 - 49CH104

Assign the oxidation states for all elements in

H2O2

+1 -1

+1

+2 -2

-1

Oxidation states

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REDOX reactions

2 H2 + O2 2 H2O

Oxidation state of H is 0Oxidation state of H is 0

Oxidation state of O is 0Oxidation state of O is 0

Oxidation state of H is 1+Oxidation state of H is 1+

Oxidation state of O is 2-Oxidation state of O is 2-

Hydrogen is oxidized and is a reducing agent.

Oxygen is reduced and is an oxidizing agent.

Lose 1 e- = LEO

Gain 2 e- = GER

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Find the oxidation state for all elements in:

HNO3

Nitrogen must be

Hydrogen is 1+Oxygen is 2-

2- 2-

5+

+1 +5 -6 = 0

Start with whatWe know

Start with whatWe know

Oxidation states

5 - 52CH104

NaCl+1

+3

-2

NaClO

Oxidation states of Cl

NaClO4

NaClO2

NaClO3

+1

+1

+1

+1

-2-2

-2-2-2

-2-2-2-2

+1

+5

+7

-1

5 - 53CH104

Types of chemical reactions

Chemical ReactionsChemical Reactions

NonredoxNonredox

Doublereplacement

Doublereplacement

CombinationCombination

RedoxRedox

Singlereplacement

Singlereplacement

CombinationCombination

Decomposition Decomposition

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Combination reactionsB + C A

REDOX or NONREDOX types

Non-REDOX reaction.

Formation of Acid Rain

SO3 + H2O H2SO4

-2+6-2-2-2-2

-2-2-2

+1+1

+1+1

+6

5 - 55CH104

Combination reactionsB + C A

REDOX or NONREDOX types

Rusting of Iron

4 Fe + 3 O2 2 Fe2O3

0-2-2-2

0 +3+3

REDOX reaction.Fe goes from 0 to +3 LEOO goes from 0 to -2 GER

5 - 56CH104

A B + C REDOX or NONREDOX types

Decomposition reactions

Non-REDOX reaction.

Decomposition of Egg Shells

CaCO3 CaO + CO2

-2+2 -2

-2-2-2-2

+4 +4+2

5 - 57CH104

Decomposition reactions

2 H2O2 2 H2O + O2

REDOX reaction.Some O goes from -1 to 0 (O2) LEO

Some O goes from -1 to -2 (H2O) GER

Decomposition of hydrogen peroxide

+1+1

-1-1

+1+1

-2 0

5 - 58CH104

A + BX B + AX Always REDOX

Iron Deposits on an Aluminum Pan

Al + FeCl3 Fe + AlCl3

00-1-1-1

-1-1-1

+3 +3

Single Replacement Reaction

REDOX reaction.Al goes from 0 to +3 LEOFe goes from +3 to 0 GER

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Activity series of metalspotassiumsodium

potassiumsodium

calciumcalcium

magnesiumaluminum

zincchromium

magnesiumaluminum

zincchromium

ironnickel

tinlead

ironnickel

tinlead

coppersilver

platinumgold

coppersilver

platinumgold

Hydrogen

Al + Fe+3 Fe + Al+3

Fe + H+ Fe+3 + H2

incr

easi

ng

rea

ctiv

ity

Element give e’s to ion

lower on list

Element give e’s to ion

lower on list

5 - 60CH104

AX + BY BX + AY Always non-REDOX

+1+1-1-1-1

-2-2-2

+5 +1

Double Replacement Reaction

Non -REDOX reaction.

AgNO3(aq) + NaCl(aq) AgCl(s)+ NaNO3(aq)

-1 -1 +1+5

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Ionic equationsIonic substances dissociate into ions when

dissolved in water.

Ag+ + NO3

-

AgNO3(aq) + NaCl (aq)

Certain ions join together

Others remain unchanged.

Ag+ NO3

-

Na+ Cl-

AgCl(s) + NaNO3(aq)

AgCl(s)+ Na+

+ Cl- + Na+ + NO3-

5 - 62CH104

Ionic equations

Total ionic equationAg+

+ NO3- +

Na+ + Cl- AgCl(s)+ Na+ +NO3-

NO3- and Na+ are spectator ions.

Net ionic equation

Ag+ + Cl- AgCl(s)

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1 pair =

1 dozen =

1 mole =

1 pair =

1 dozen =

1 mole =

The Mole

1 mol eggs___

6.02 x 1023eggs

1 mol Au_______

6.02 x 1023 Au atoms

_____1 mole H2O_____

6.02 x 1023 H2O molecules

2

12

6.02 x 1023

602,000,000,000,000,000,000,000.

5 - 64CH104

1 car ___ 4 wheels

The Mole & Formulas

1 mol cars_

4 mol wheels1 doz cars4 doz wheels

1 mole H2O

2 mol H

1 mole H2O

1 mol O

5 - 65CH104

= doz wheels

1 car ___ 4 wheels

The Mole & Formulas

1 mol cars_

4 mol wheels1 doz cars4 doz wheels

4 doz wheels

1doz cars

5 doz cars

120

2 mol H

1 mol H2O

5 mol H2O

1

10 = mol H

5 - 66CH104

1 mole = MW in g’s The Mole & Molar Mass

1 mol Au_

197 g Au

1 mol Au_

197 g Au

1 mole Au = 197 g Au1 mole Au = 197 g Au

197 g Au

1 mol Au

197 g Au

1 mol Au

__197 g Au _

6.02 x 1023 atoms Au

__197 g Au _

6.02 x 1023 atoms Au

5 - 67CH104

1 mole = MW in g’s The Mole & Molar Mass

1 mol S_

32 g S

1 mol S_

32 g S

1 mole S = 32 g S1 mole S = 32 g S

32 g S

1 mol S

32 g S

1 mol S

1 mol C

12 g C

1 mol C

12 g C

1 mole C = 12 g C1 mole C = 12 g C

12 g C

1 mol C

12 g C

1 mol C

5 - 68CH104

The Mole & Molar Mass 1 mol H2O_

18.0 g H2O 1 mole H2O has:

1.0 g H =

1 mol H

2 mol H

1

2.0 g H

16.0 g O =

1 mol O

1 mol O

1

16.0 g O

18.0 g H2O

1 mol H2O

18.0 g

5 - 69CH104

Molar MassFind the MW of Glucose; C6H12O6

1.0 g H =

1 mol H

12 mol H

1

12.0 g H

16.0 g O =

1 mol O

6 mol O

1

96.0 g O

180.0 g C6H12O6

1 mol C6H12O6

12.0 g C =

1 mol C

6 mol C

172.0 g C

5 - 70CH104

1 mol H2O =

18 g H2O

Mass to Mole Conversions

How many moles of water are in 36 g H2O?

What should the answer look like?

What is Unique to the problem?

36 g H2O

1

mol H2O2.0

5 - 71CH104

1 mol H2O

18 g H2O

Mass to Mole Conversions

How many moles of H are in 36 g H2O?

What should the answer look like?

What is Unique to the problem?

36 g H2O

1

mol H4.0 2 mol H =

1 mol H2O

5 - 72CH104

180 g Gluc =

1 mol Gluc

Mole to Mass ConversionsHow many g’s of Glucose (C6H12O6)

are in 5 mol Glucose?What should the answer look like?

What is Unique to the problem?

5 mol Gluc

1g Glucose900

5 - 73CH104

You need a balancedequation.

You need a balancedequation.

H2 + O2 -----> H2O H2 + O2 -----> H2O

The mole and chemical equations

Stoichiometry - Calculations of quantities in a chemical rxn.

2 2

5 - 74CH104

Moles in Chemical Equations

2 Na(s) + Cl2 (g) 2 NaCl

2 mol Na

1 mol Cl2

2 mol Na

2 mol NaCl

1 mol Cl2

2 mol NaCl

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Moles in Chemical Equations2 Na(s) + Cl2 (g) 2 NaCl

2 mol Na

1 mol Cl2

2 mol Na

2 mol NaCl

1 mol Cl2

2 mol NaCl

How many moles of Cl2 are needed to completely react with 40 moles of Na?How many moles of Cl2 are needed to completely react with 40 moles of Na?

1 mol Cl2 =

2 mol Na

What is Unique to the problem?

40 mol Na 1

20 mol Cl2

What should the answer look like?

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1mol Na23.0g Na

2 Na(s) + Cl2 (g) 2 NaCl

1 mol Cl2

2 mol Na

What is Unique to the problem?

40 g Na1

61.7 g Cl2

What should the answer look like?

g’s Cl2

Mass in Chemical Equations

mols Na mols Cl2

70.9 g Cl2 =

1 mol Cl2

How many g’s of Cl2 are needed to completely react with 40 g’s of Na?How many g’s of Cl2 are needed to completely react with 40 g’s of Na?

40 g’s Na

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1mol Fe2O3

159.7g Fe2O3

Fe2O3 + 3H2 2Fe + 3H2O

2 mol Fe1 mol Fe2O3

What is Unique to the problem?

20.g Fe2O3

114 g Fe

What should the answer look like?

g’s FeMass in Chemical Equations

mols Fe2O3 mols Fe

55.8 g Fe = 1 mol Fe

How many g’s of Fe can be produced from 20. g’s of Fe2O3?

How many g’s of Fe can be produced from 20. g’s of Fe2O3?

20.g’s Fe2O3

5 - 78CH104

1mol Fe2O3

159.7g Fe2O3

Fe2O3 + 3H2 2Fe + 3H2O

2 mol Fe1 mol Fe2O3

20.g Fe2O3

1

14 g Fe

Percent Yield

55.8 g Fe = 1 mol Fe

How many g Fe can be made from 20.g Fe2O3?How many g Fe can be made from 20.g Fe2O3?

What is the percent yield if I only got 12 g’s of Fe from 20. g’s of Fe2O3?

What is the percent yield if I only got 12 g’s of Fe from 20. g’s of Fe2O3?

% yield = Actual x 100 Theoretical% yield = Actual x 100 Theoretical

% yield = 12 g x 100 = 14 g% yield = 12 g x 100 = 14 g

85.7 %85.7 %

5 - 79CH104

Limiting reactant1 c milk + 2 c bisquick + 1 Tbs oil +1 egg = 18

pancakes

If I have 5 c. Bisquick and 2 eggs, how many pancakes can I make?

From 5 c. Bisquick

From 2 eggs

5 c. Bquick

2 eggs

18 p cakes =

2 c. Bquick

45 p cakes

36 p cakes18 p cakes =

1 egg

5 - 80CH104

Limiting reactant1 c milk + 2 c bisquick + 1 Tbs oil +1 egg = 18

pancakes

If I have 5 c. Bisquick and 2 eggs, how many pancakes can I make?

From 5 c. Bisquick

From 2 eggs

45 p cakes

36 p cakes

So: Eggs are the limiting reagent.I can make only 36 pancakes I’ll have bisquick left over.

5 - 81CH104

Limiting reactant

If I have 20.g of Fe2O3 and 2.0g H2,

how many g’s Fe can I make?

From 20. g Fe2O3

From 2.0 g H2

14 g Fe

Fe2O3 + 3H2 2Fe + 3H2O

1mol H2

2.02 g H2

2 mol Fe3 mol H2

2.g H2

137 g Fe 55.8 g Fe =

1 mol Fe

So: Fe2O3 is the limiting reagent.

I can make only 14 g FeI’ll have H2 left over.

5 - 82CH104

31.7 18 g H2O =

1 molH2O

13.5 18 g H2O =

1 molH2O

Limiting reactant

If I have 20.g of NH3 and 20.g O2,

how many g’s H2O can I make?From 20. g NH3

From 20. g O2

4NH3 + 5O2 4NO + 6H2O

1mol NH3

17 g NH3

6 mol H2O

4 mol NH3

20.g NH3

1g H2O

1mol O2

32 g O2

6 mol H2O

5 mol O2

20.g O2

1

g H2O

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Limiting reactant

If I have 20.g of NH3 and 20.g O2,

how many g’s H2O can I make?

From 20. g NH3

From 20. g O2

31.7 g H2O

4NH3 + 5O2 4NO + 6H2O

So: O2 is the limiting reagent.

We can make only 13.5 g H2OWe’ll have NH3 left over.

13.5 g H2O

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13 76 g CS2 =

1 mol CS2

5.9 76 g CS2 =

1 mol CS2

Limiting reactant

If I have 10.g of C and 10.g SO2,

how many g’s CS2 can I make?From 10. g C

From 10. g SO2

5C + 2SO2 1CS2 + 4CO

1mol C

12 g C

1 mol CS2

5 mol C

10.g C1

g CS2

1mol SO2

64 g SO2

1 mol CS2

2 mol SO2

10.g SO2

1

g CS2

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Limiting reactant

13 g CS2

So: SO2 is the limiting reagent.

We can make only 5.9 g CS2

We’ll have C left over.

5.9 g CS2

If I have 10.g of C and 10.g SO2,

how many g’s CS2 can I make?From 10. g C

From 10. g SO2

5C + 2SO2 1CS2 + 4CO

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Study of Energy changes in Reactions =

• used to calculate the amount of useful work produced by chemical reactions.

Thermodynamics

Enthalpy

andEntropy

= Change in Energy = DH

= Change in = DS

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law of Thermodynamics

“Energy can’t be created or destroyed in a chemical reaction”

• Energy just gets

converted from one form to another.

(you can’t get something from nothing)

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Exothermic Reactions

Mg + O2 MgO + Energy

CH4 + 2O2 CO2 + 2H2O + 211 kcal

Breaking bondscosts E(+DH)

Making bondsgives E(-DH)

Old bonds break New bonds get made

Hproducts - Hreactants = H

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Activation Energy (Eact)The minimum amount of energy required to produce a chemical reaction.

• The energy of collision must be great enough to break the old bonds and form the new ones.

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H2OH2O

more stablemore stable

Energy in Chemical Reactions

Exothermic reactionExothermic reaction

-H= heat of reaction

En

ergy

Rxn Progress

Reactants

Products

Eact= Activation Energy

(Gets hot)(Gets hot)

H2 + O2H2 + O2

2H2 + O22H2 + O2 2H2O + Energy2H2O + Energy

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Exothermic (Exergonic) RxnsE

ner

gy

Rxn Progress

2Mg + O2 2MgO + Energy

2H2 + O2 2H2O + Energy

CH4 + 2O2 CO2 + 2H2O + 213 kcal

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Exothermic ReactionsE

ner

gy

Rxn Progress

Reactants(Water)

Products(Ice)

Energy is released

Products are more stable.

-DH

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more stablemore stable

Energy in Chemical Reactions

Endothermic reactionEndothermic reaction

+H= heat of reactionEn

ergy

Rxn Progress

Reactants

Products

Eact= Activation Energy

(Gets cold)(Gets cold)

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En

ergy

Rxn Progress

Endothermic (Endergonic) RxnsEnergy is required

Products are less stable.

Products(Water)

Reactants(Ice)

+DH

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Examples of Energy diagrams

Exothermic reaction

Endothermic reaction

H > 0reactants are more stable

H < 0products are more stable

+H

-H

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Examples ofenergy diagrams

High activation energyLow heat of reaction

Low activation energyHigh heat of reaction

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Energy in Reactions

How many kilojoules are released when 100. g Na reacts with chlorine?

2Na(s) + Cl2(g) 2NaCl + 819 kJ

1mol Na

23.0 g Na

819 kJ 2 mol Na

100.g Na1

= kJ