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1 5.1  Introduction to Chemical Reactions 1 5.1  Introduction to Chemical Reactions

1 5.1 Introduction to Chemical Reactions - PowerPoint Presentation

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1 5.1 Introduction to Chemical Reactions - PPT Presentation

A General Features of Physical and Chemical Changes A chemical change a chemical reaction converts one substance into another Chemical reactions involve Breaking bonds in the reactants ID: 790981

mass chemical mol grams chemical mass grams mol mole number moles h2o molecules reactant reaction atoms equations step formed

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Slide1

1

5.1 Introduction to Chemical ReactionsA. General Features of Physical and Chemical Changes

A chemical change (a chemical reaction) converts one substance into another.

Chemical reactions involve:

Breaking bonds in the reactants (starting materials)

Forming new bonds in the products

Slide2

2

5.1 Introduction to Chemical ReactionsA. General Features of Physical and Chemical Changes

Slide3

3

5.1 Introduction to Chemical ReactionsB. Writing Chemical Equations

Slide4

4

5.1 Introduction to Chemical ReactionsB. Writing Chemical Equations

A chemical equation uses chemical formulas and othersymbols showing what reactants are the starting

materials in a reaction and what products are formed.

Slide5

5

5.1 Introduction to Chemical ReactionsB. Writing Chemical Equations

The law of conservation of mass states that atoms cannot be created or destroyed in a chemical reaction.

Slide6

6

5.2 Balancing Chemical Equations

HOW TO Balance a Chemical EquationExample

Write a balanced chemical equation forthe reaction of propane (C3H8) with oxygen (O2) to form carbon dioxide (CO2)and water (H2O).Step [1]

Write the equation with the correct formulas.

Slide7

7

5.2 Balancing Chemical Equations

Step [2]HOW TO Balance a Chemical Equation

Balance the equation with coefficients oneelement at a time.

Balance the C’s first:

Balance the H’s next:

Slide8

8

5.2 Balancing Chemical Equations

Step [2]HOW TO Balance a Chemical Equation

Balance the equation with coefficients oneelement at a time.

Finally, balance the O’s:

Slide9

Balance the Equations9__H2 + __O2  __H2O__NO + __O2  __NO2__CH4 + __Cl2  __CH

2Cl2 + __HCl

Slide10

Polyatomic Ions10__Ca3(PO4)2 + __H2SO4  __CaSO4 + __H3PO4

Slide11

Balance the Equation11__Al + __H2SO4  __Al2(SO4)3 + __H2__Na2SO3 + __H3PO4

 __H2SO3 + __Na

3PO4

Slide12

Balance the Equation12__Mg + __HBr  __MgBr2 + __H2__KClO3  __KCl + __O2__CH4 + __Cl2  __CCl4 + __HCl

Slide13

Balance the Equation13__Al2O3 + __HCl  __AlCl3 + __H2O__Al(OH)3 + __H2SO4  __Al2(SO4

)3 + __H2O

Slide14

__Ni + __HCl  __NiCl2 + __H2__PbS + __O2  __PbO + __SO2__H3PO4 + __Ca(OH)2  __Ca3(PO4

)2 + __H2

O

Slide15

__H2SO4 + __NaOH  __Na2SO4 + __H2O__CO + __O2  __CO2__S + __O2 + __H2

O  __H2SO4

Slide16

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5.3 Types of Reactions

The majority of chemical reactions fall into 6 categories:

combinationdecompositionsingle replacement

double replacementoxidation and reduction (Section 5.4)acid-base (Chapter 9)

Slide17

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5.3 Types of ReactionsA. Combination and Decomposition

A combination reaction is the joining of two or more reactants to form a single product.

Slide18

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5.3 Types of ReactionsA. Combination and Decomposition

A decomposition reaction is the conversion of a single reactant to two or more products.

Slide19

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5.3 Types of ReactionsB. Replacement Reactions

A single replacement reaction is a reaction in which one element replaces another element in a compound to form a different compound and element as products.

Slide20

205.3 Types of ReactionsB. Replacement Reactions2 NaCl + Br2 Fe + CaSO4 

Slide21

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5.3 Types of ReactionsB. Replacement Reactions

A double replacement reaction is a reaction in which two compounds exchange “parts”–atoms or ions—to form two new compounds.

Slide22

225.3 Types of ReactionsB. Replacement ReactionsAgNO3 + NaClHCl + NaOH

Slide23

Combination, Decomposition, Single Displacement or Double Displacement?23Ni(NO3)2 + Mg  Ni + Mg(NO3)22 KI + Sn(NO3)2  SnI2 + 2 KNO3

2 HgO

 2 Hg + O2

Slide24

24Mg + 2 ZnCl  MgCl2 + 2 ZnH2C=CH2 + HBr  CH3CH2BrKCN + HCl 

KCl + HCN

Slide25

256 Na + AlSO4  3 NaSO4 + 2 AlKNO3 + HBr  KBr + HNO3

Slide26

Predicting Reactions26Combination: N2 + ____  Mg3N2Decomposition: 2 SO3  2SO2 + ______Single Replacement: 2 Ag + CuBr2  ______ + ______Double Replacement: KOH + HI  ______ + _______

Slide27

27Comb: 2 Na + Cl2 Decomp: 2 NI Single: Cl2 + Ki Double: NaOH + HBr 

Slide28

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5.4 Oxidation and ReductionA. General Features

Oxidation is the loss of electrons from an atom.

Reduction

is the gain of electrons by an atom.

Both processes occur together in a single reaction called an oxidation−reduction or redox reaction.

A

redox

reaction involves the

transfer of electrons

from one element to another.

A redox reaction always has two components, one that is oxidized and one that is reduced.

Slide29

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5.4 Oxidation and ReductionA. General Features

Slide30

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5.4 Oxidation and ReductionA. General Features

Oxidation half reaction:

Each of these processes can be written as an individual half reaction

:Zn + Cu2+

Zn

2+

+ Cu

Reduction

half reaction:

Slide31

31

5.4 Oxidation and ReductionA. General Features

Zn + Cu2+

Zn

2+ + Cu

A compound that is reduced while causing anothercompound to be oxidized is called an oxidizing agent.

Cu

2+

acts as an

oxidizing agent

because it causes

Zn to lose electrons and become

oxidized

.

oxidized

reduced

Slide32

32

5.4 Oxidation and ReductionA. General Features

Zn + Cu2+

Zn

2+ + Cu

Zn acts as a reducing agent because it causes Cu2+ to gain electrons and become reduced.

A compound that is

oxidized

while causing another

compound to be

reduced

is called a

reducing agent

.

oxidized

reduced

Slide33

33

5.4 Oxidation and ReductionA. General Features

Slide34

34

5.4 Oxidation and Reduction

B. Examples of Oxidation–Reduction Reactions

Iron Rusting

4 Fe(s) + 3 O2(g)

2 Fe2O3(s)

Fe

3+

O

2–

neutral Fe

neutral O

Fe loses e

and is oxidized.

O gains e

and is reduced.

Slide35

35

5.4 Oxidation and ReductionB. Examples of Oxidation–Reduction Reactions

Zn + 2 MnO2

ZnO + Mn

2O3

Slide36

Zn + 2H+  Zn2+ + H2Fe3+ + Al  Al3+ + Fe

Slide37

I- + Br2  I2 + Br-AgBr  Ag + Br2

Slide38

38

5.4 Oxidation and ReductionB. Examples of Oxidation–Reduction Reactions

Oxidation results in the:

Reduction results in the:

Gain of oxygen atoms

Loss of hydrogen atoms

Loss

of

oxygen

atoms

Gain

of

hydrogen

atoms

Slide39

39

5.5 The Mole and Avogadro’s Number

A mole is a quantity that contains 6.02 x 1023 items.

1 mole of C

atoms = 6.02 x 1023 C atoms

1 mole of H2O molecules = 6.02 x 1023 H2O molecules

1 mole of Vitamin C

molecules

= 6.02 x 10

23

Vitamin C

molecules

The number 6.02 x 10

23

is

Avogadro’

s number

.

1.3×10

23

 kg

Titan

, largest moon of Saturn

1.5×10

23

 kg

Ganymede

, largest moon of Jupiter

3.3×10

23

 kg

Mercury

6.4×10

23

 kg

Mars

Slide40

40

How many items do 1 mol of the following contain:

BaseballsBicycles

CheeriosCH4 molecules

Slide41

41

5.5 The Mole and Avogadro’s Number

It can be used as a conversion factor to relate thenumber of moles

of a substance to the number ofatoms or molecules:

1 mol6.02 x 1023 atoms

or

6.02 x 10

23

atoms

1 mol

1 mol

6.02 x 10

23

molecules

or

6.02 x 10

23

molecules

1 mol

Slide42

42

5.5 The Mole and Avogadro’s Number

Sample Problem 5.5

How many molecules are contained in 5.0 moles of carbon dioxide (CO2)?Step [1]

Identify the original quantity and the desired quantity.

Slide43

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5.5 The Mole and Avogadro’s Number

Step [3]

Set up and solve the problem.Step [2]

Write out the conversion factors.

Slide44

44

How many C atoms are there in the following:

2.0 mol6.0 mol0.5 mol25.0 mol

Slide45

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How many molecules are contained in each of the following number of moles

2.5mol of penicillin0.25 mol of NH30.4 mol of Sugar55.3 mol of Acetaminophen

Slide46

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5.6 Mass to Mole Conversions

The formula weight is the sum of the atomic weights of all the atoms in a compound, reported in atomic mass units (amu).

HOW TO Calculate the Formula Weight of a CompoundExample

Calculate the formula weight for FeSO4.Step [1]

Write the correct formula and determinethe number of atoms of each element fromthe subscripts.

Slide47

47

5.6 Mass to Mole Conversions

HOW TO Calculate the Formula Weight of a CompoundStep [2]

Multiply the number of atoms of each element by the atomic weight and addthe results.

Slide48

48

5.6 Mass to Mole ConversionsA. Molar Mass

The molar mass is the mass of one mole of any substance, reported in grams per mole (g/mol).

The value of the molar mass of a compound in grams equals the value of its formula weight in amu.

Slide49

49

5.6 Mass to Mole ConversionsB. Relating Grams to Moles

The molar mass relates the number of moles to the number of

grams of a substance.

In this way, molar mass can be used as a conversion factor.

The molar mass of H2O is 18.0 g/mol, the conversion factor can be written:

Slide50

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5.6 Mass to Mole ConversionsB. Relating Grams to Moles

Sample Problem 5.9

How many moles are present in 100. g of aspirin (C9H8O4)?Step [1]

Calculate the molar mass.

Slide51

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5.6 Mass to Mole ConversionsB. Relating Grams to Moles

Step [2]

Write out the conversion factors.

The conversion factor is the molar mass, and it can be written in two ways.

Choose the one that places the unwanted unit, grams, in the denominator so that the units cancel:

Slide52

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5.6 Mass to Mole ConversionsB. Relating Grams to Moles

Step [3]

Set up and solve the problem.

Slide53

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How many moles are contained in the following:

100 g NaCl

Slide54

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How many moles are contained in the following:

0.25g Aspirin(C9H8O4)

Slide55

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How many moles are contained in the following:

25.5g CH4

Slide56

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How many moles are contained in the following:

25g of H2O

Slide57

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5.6 Mass to Mole ConversionsC. Relating Grams to Number of Atoms or Molecules

We can also use the molar mass to show the relationship between grams and number of

molecules (or atoms).

Slide58

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5.6 Mass to Mole ConversionsC. Relating Grams to Number of Atoms or Molecules

Sample Problem 5.10

How many molecules are in a 325-mg tablet of aspirin (C9H8O4)?Step [1]

Find the molar mass

Slide59

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5.6 Mass to Mole ConversionsC. Relating Grams to Number of Atoms or Molecules

Step [2]

Write out the conversion factors.

Slide60

60

5.6 Mass to Mole ConversionsC. Relating Grams to Number of Atoms or Molecules

Step [3]

Set up and solve the problem.

Slide61

How many molecules are present in a 500mg tablet of penicillin (C16H18N2O4S)61

Slide62

How many molecules are present in a 500mg tablet of penicillin (C16H18N2O4S)62

Slide63

How many molecules are present in a 750mg of Mescaline (Hallucinogenic from peyote) (C11H17NO3)63

Slide64

64

5.7 Mole Calculations in Chemical Equations

A balanced chemical equation also tell us:

The number of moles of each reactant that combineThe number of moles of each product formed

1 N2(g

)+

1 O

2

(

g

)

2 NO(

g

)

(The coefficient

1

has been written for emphasis.)

Slide65

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5.7 Mole Calculations in Chemical Equations

Coefficients are used to form mole ratios, which canserve as

conversion factors.

N2(g)

+ O2(g)

2 NO(

g

)

Mole ratios:

Slide66

66

5.7 Mole Calculations in Chemical Equations

Sample Problem 5.11

Using the balanced chemical equation, how many moles of CO are produced from 3.5 molesof C2H6?Step [1]

Identify the original and desired quantities.

2 C2H6(g) + 5 O2(g)

4 CO(

g

) + 6 H

2

O(

g

)

Slide67

67

5.7 Mole Calculations in Chemical Equations

Step [2]

Write out the conversion factors.Step [3]

Set up and solve the problem.

2 C2H6(g) + 5 O2(g)

4 CO(

g

) + 6 H

2

O(

g

)

Slide68

Complete the following conversionsHow many mol of NO are formed from 3.3 mol of N2How many mol of NO are formed from 0.5mol of O2How many moles of O2 are needed to completely react with 1.2 mol of N268

N

2(g)

+

O2(g)

2 NO(g)

Slide69

How many mol of NO are formed from 3.3 mol of N2How many mol of NO are formed from 0.5mol of O269

N

2(g)

+

O2(g)

2 NO(g)

Slide70

How many moles of O2 are needed to completely react with 1.2 mol of N270

N

2(g)

+

O2(g)

2 NO(g)

Slide71

71

5.8 Mass Calculations in Chemical Equations

HOW TO Convert Moles of Reactant to Grams of ProductExample

Using the balanced equation, how manygrams of O3 are formed from 9.0 mol of O2.

3 O2(g)

2 O3(g)

sunlight

Slide72

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5.8 Mass Calculations in Chemical Equations

Step [1]

Convert the number of moles of reactantto the number of moles of product usinga mole–mole conversion factor.HOW TO Convert Moles of Reactant to Grams of Product

3 O2(g)

2 O3(g)

sunlight

Slide73

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5.8 Mass Calculations in Chemical Equations

HOW TO Convert Moles of Reactant to Grams of ProductStep [2]

Convert the number of moles of productto the number of grams of product using the product’s molar mass.

Slide74

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5.8 Mass Calculations in Chemical Equations

HOW TO Convert Moles of Reactant to Grams of Product

Set up and solve the conversion.

3 O2(g)

2 O3(g)

sunlight

Slide75

C6H12O6 2 C2H6O + 2CO2How many g of ethanol are formed from 0.55 mol of glucoseHow many g of CO2 are formed from 0.25 mol of glucoseHow many g of glucose are needed to form 1mol of ethanol75 (ethanol) (glucose)

Slide76

How many g of ethanol are formed from 0.55 mol of glucose76

C6

H12O6 2 C2H6O + 2CO2

Slide77

How many g of CO2 are formed from 0.25 mol of glucose77

C

6H12O6 2 C2H6O + 2CO2

Slide78

How many g of ethanol would be formed if 0.55 mols of glucose is used78

C6

H12O6 2 C2H6O + 2CO2

Slide79

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5.8 Mass Calculations in Chemical Equations

HOW TO Convert Grams of Reactant to Grams of ProductExample

Ethanol (C2H6O) is synthesized by reacting ethylene (C2H4) with water.How many grams of ethanol are formed from 14 g of ethylene?

C2H4 + H2O

C2H6O

Slide80

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5.8 Mass Calculations in Chemical Equations

HOW TO Convert Grams of Reactant to Grams of Product

C2H4 + H2O

C2H6O

Slide81

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5.8 Mass Calculations in Chemical Equations

HOW TO Convert Grams of Reactant to Grams of Product

C2H4 + H2O

C2H6O

Slide82

82

5.8 Mass Calculations in Chemical Equations

HOW TO Convert Grams of Reactant to Grams of Product

C2H4 + H2O

C2H6O

Slide83

83

Slide84

1) How many molecules of CO is produced?2) How many g of C2H6 are needed to react with all of O23) How many g of H2O are produced when 15g of O2 is used.84

2 C

2H6(g) + 5 O2(g)

4 CO(g) + 6 H2O(g)

Slide85

2 C

2H6(g

) + 5 O2(g)

4 CO(g) + 6 H2O(g)How many molecules of CO is produced?

Slide86

2 C

2H6(g

) + 5 O2(g)

4 CO(g) + 6 H2O(g)How many g of C2H6 are needed to react with all of O2

Slide87

2 C

2H6(g

) + 5 O2(g)

4 CO(g) + 6 H2O(g)How many g of H2O are produced when 15g of O2 is used.

Slide88

2 NO + O2  2 NO2How many g of NO2 are formed from 20g of NO?

Slide89

2 NO + O2  2 NO2How many g of NO are needed to make 30g of NO2?

Slide90

2 CH4O + 3 O2  2 CO2 + 4 H2OHow many grams of CO2 are formed from 50g CH4O?

Slide91

91

5.9 Percent Yield

The theoretical yield is the amount of product expected from a given amount of reactant based on the coefficients in the balanced chemical equation.

The percent yield is the amount of product isolated from a reaction.

Usually, however, the amount of product formed is less than the maximum amount of product predicted.

Slide92

92

5.9 Percent Yield

Sample Problem 5.14

If the reaction of ethylene with water to form ethanol has a calculated theoretical yield of 23 g of ethanol, what is the percent yield if only 15 g of ethanol are actually formed?

Slide93

93

5.10 Limiting Reactants

The limiting reactant is the reactant that is completely used up in a reaction.

Slide94

94

5.10 Limiting ReactantA. Determining the Limiting Reactant

Analyze the two possible outcomes:

If the amount present of the second reactantis

less than what is needed, the secondreactant is the limiting reagent.

If the amount present of the second reactant isgreater than what is needed, the secondreactant is in excess.

Slide95

95

5.10 Limiting ReactantC. Determining the Limiting Reactant Using the Number of Grams

Sample Problem 5.20

Using the balanced equation, determine the limitingreactant when 10.0 g of N2 (MM = 28.02 g/mol) reactwith 10.0 g of O2 (MM = 32.00 g/mol).

N2(g) + O2(g) 2 NO(g)

Slide96

96

5.10 Limiting ReactantC. Determining the Limiting Reactant Using the Number of Grams

Sample Problem 5.20

[1] Convert the number of grams of each reactant into moles using the molar masses.

Slide97

97

5.10 Limiting ReactantC. Determining the Limiting Reactant Using the Number of Grams

Sample Problem 5.20

[2] Determine the limiting reactant by choosing N2 as the original quantity and converting to mol O2.

1 mol O21 mol N2

0.357 mol N2 x

= 0.357 mol O

2

mole–mole

Conversion factor

The amount of O

2

we started with (0.313 mol) is

less than the amount we would need (0.357 mol) so

O

2

is the limiting reagent

.

Slide98

N2(g) + O2(g)  NO(g)Determine the limiting reactant under the following conditions:1) 12.5g N2 and 15.0g O22) 14.0g N2 and 13.0g O298

Slide99

12.5g N2 and 15.0g O2N2(g) + O2(g)  NO(g)

Slide100

N2(g) + O2(g)  NO(g)14.0g N2 and 13.0g O2

Slide101

101Balance/Rxn Type/ REDOXNi + HCl  NiCl2 + H2CH4 + Cl2  CCl4 + HClKClO3  KCl + O2

Al2O3

+ HCl  AlCl3 + H2O

Slide102

102

Slide103

103

Slide104

104

Slide105

105Given the following reaction:C12H22O11 + H2O  C2H6O + CO2 (sucrose) (ethanol) Balance the equationDetermine the molecular weight of sucroseHow many mols of Ethanol would be produced from 2

mols of sucrose?

How many g of ethanol would be produced from 17.1 g sucrose?

Slide106

106