chapter 5 chemical quantities and reactions

Chemistry An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc.

Chapter 5

Chemical Quantities and Reactions

1

Chemistry for Health Sciences

Prof. Ralph Duarte


The Mole

Counting Units

Counting units

A counting term states a specific number of items.

1 dozen eggs = 12 eggs

1 case soda = 24 cans

1 ream = 500 sheets of paper

2


Avogadro's Number

Small particles such as atoms, formula units, molecules, and ions are counted using the mole.

1 mole = 6.022 x 1023 items

Avogadro’s number

602 200 000 000 000 000 000 000 = 6.022 x 1023

3


Mole of Atoms

1 mole of an element = 6.022 x 1023 atoms of that element

1 mole of carbon = 6.022 x 1023 atoms of carbon

1 mole of sodium = 6.022 x 1023 atoms of sodium

4


Number of Particles in One-Mole Samples

5


Avogadro's Number

Avogadro’s number, 6.022 x 1023, can be written as an

equality and as two conversion factors.

Equality:

1 mole = 6.022 x 1023 particles

That means we can use the following Conversion Factors as needed:

6.022 x 1023 particles or 1 mole 1 mole 6.022 x 1023 particles

6


Converting Moles to Particles

(Atoms, formula units or molecules)

Step 1 State the data and question:

Data: 0.50 mole of CO2

Question: molecules of CO2

7

Problem:

How many CO2 molecules are in 0.50 mole of CO2?

Avogadro’s number is used to convert moles of a substance to particles.



(Atoms, formula units or molecules) 8

Question = Data x “question” units

“data” units

Include conversion factor here

Problem:





molecules

mole

Problem:


Use Avogadro’s number to write conversion factors:

1 mole of CO2 = 6.022 x 1023 molecules of CO2

6.022 x 1023 CO2 molecules or 1 mole CO2

1 mole CO2 6.022 x 1023 CO2 molecules

CO2 molecules = 0.50 mole CO2 x




6.022 x 1023 CO2 molecules

1 mole CO2

Problem:



1 mole of CO2 = 6.022 x 1023 molecules of CO2

6.022 x 1023 CO2 molecules or 1 mole CO2

1 mole CO2 6.022 x 1023 CO2 molecules

CO2 molecules = 0.50 mole CO2 x




CO2 molecules = 0.50 mole CO2 x 6.022 x 1023 CO2 molecules

1 mole CO2

Problem:


= 3.0 x 1023 molecules of CO2


Learning Check

The number of atoms in 2.0 mole of Al atoms is:

A. 2.0 Al atoms

B. 3.0 x 1023 Al atoms

C. 1.2 x 1024 Al atoms

12


Solution

The number of atoms in 2.0 moles of Al atoms is:

= 2.0 moles Al x 6.022 x 1023 Al atoms

1 mole Al = 12.044 x 1023 Al atoms = 1.2 x 1024 Al atoms

13

The number of atoms in 2.0 mole of Al atoms is:

A. 2.0 Al atoms

B. 3.0 x 1023 Al atoms

C. 1.2 x 1024 Al atoms


Learning Check

Calculate the number of moles of S in 1.8 x 1024 atoms of S

14


Solution

Set up the problem to calculate the number of moles of S:

= 1.8 x 1024 S atoms x 1 mole S

6.022 x 1023 S atoms = 0.29890 moles x 1024 x 10-23

= 3.0 moles of S atoms

15

Calculate the number of moles of S in 1.8 x 1024 atoms of S

State the needed and given quantities:

Given: 1.8 x 1024 atoms of S

Needed: moles of S


Learning Check

How many molecules of aspirin are in 0.150 mole of aspirin, C9H8O4?

16




molecules

mole


1 mole of C9H8O4 = 6.022 x 1023 molecules of C9H8O4

6.022 x 1023 molecules or 1 mole

1 mole 6.022 x 1023 molecules

molecules = 0.150 mole x



Solution

Set up the problem to calculate the number of particles.

= 0.150 mole C9H8O4 x 6.022 x 1023 molecules C9H8O4

1 mole C9H8O4

= 0.903 x 1023 molecules C9H8O4

= 9.03 x 10-1 x 1023 molecules C9H8O4

= 9.03 x 1022 molecules C9H8O4

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One-Mole Quantities

19

1 mole S = 32.07 g 1 mole Fe = 55.85 g 1 mole NaCl = 58.44 g 1 mole K2Cr2O7 = 294.2 g 1 mole C12H22O11 = 342.3 g

1 mole of S

= 6.022 x 1023 atoms of S

1 mole of Fe

= 6.022 x 1023 atoms of Fe

1 mole of NaCl = 6.022 x 1023 Formula Units

of NaCl

1 mole of K2Cr2O7 = 6.022 x 1023 Formula Units

of K2Cr2O7

1 mole of C12H22O11 = 6.022 x 1023 molecules

of C12H22O11

Potassium

Dichromate Sucrose


Molar Mass The molar mass • is the mass of 1 mole of an element

• is the atomic mass expressed in grams

20


Learning Check

Give the molar mass for each element:

A. 1 mole of K atoms = ________

B. 1 mole of Sn atoms = ________

C. 1 mole of Ca atoms = ________

21


Solution

Give the molar mass for each element:

A. 1 mole of K atoms = 39.10 g of K

B. 1 mole of Sn atoms = 118.7 g of Sn

C. 1 mole of Ca atoms = 40.08 g of Ca

22


Molar Mass of the Compound

Example: Ethanol, C2H6O

To calculate the molar mass of Ethanol, C2H6O:

Step 1 Obtain the molar mass of each element.

molar mass of C = 12.01 g of C

molar mass of H = 1.008 g of H

molar mass of O = 16.00 g of O

Step 2 Multiply each molar mass by the number of subscripts in the formula

23

Step 3 Calculate the molar mass by adding the masses of the elements:

2 x molar mass of C = 24.02 g of C

6 x molar mass of H = 6.05 g of H

1 x molar mass of O = 16.00 g of O

Molar mass of C2H6O = 46.07 g of C2H6O


Learning Check

What is the molar mass of each compound?

A. K2O

B. Al(OH)3

24


Solution

What is the molar mass of each compound?

Calculate the molar mass by adding the molar masses of the elements:

K2O 2 x molar mass of K = 78.20 g of C


Molar mass of K2O = 94.20 g of K2O

Al(OH)3 1 x molar mass of Al = 26.98 g of Al


3 x molar mass of H = 3.024 g of H Molar mass of Al(OH)3 = 78.00 g of Al(OH)3

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Calculations Using Molar Mass

Molar mass conversion factors

• are fractions (ratios) written from the molar mass.

• relate grams and moles of an element or compound.

• for methane, CH4, used in gas stoves and gas heaters, is

1 mole of CH4 = 16.0 g of CH4 (molar mass equality)

Conversion factors:

16.0 g CH4 or 1 mole CH4

1 mole CH4 16.0 g CH4

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Converting Mass to Moles of a Compound

A box of table salt, NaCl, contains 737 g of NaCl.

How many moles of NaCl are in the box?

27


Converting Mass to Moles of

Compound NaCl

Step 1 State the data and question. Data: 737 g of NaCl

Question: moles of NaCl

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A box of table salt, NaCl, contains 737 g of NaCl.

How many moles of NaCl are in the box?

Step 2 Set up the problem to convert grams to moles.

moles NaCl = 737 g NaCl x 1 mole NaCl

58.44 g NaCl

= 12.6 moles NaCl


Map: Mass – Moles – Particles

29


Learning Check

Allyl sulfide, C6H10S, is a compound that has

the odor of garlic. How many moles of

C6H10S are in 225 g?

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Solution

Step 1 State the data and question. Data: 225 g of C6H10S

Question: moles of C6H10S

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Step 2 Determine the molar mass and write conversion factors.

Conversion Factors 1 mole of C6H10S = molar mass of C6H10S = 114.2 g of C6H10S 114.2 g C6H10S or 1 mole C6H10S 1 mole C6H10S 114.2 g C6H10S

Step 3 Set up the problem to convert grams to moles. mole of C6H10S = 225 g C6H10S x 1 mole C6H10S 114.2 g C6H10S

= 1.97 moles of C6H10S


Chemical Changes

A chemical change

• occurs when a substance is converted into one or more

substances with different formulas and different properties

• may be observed by the formation of bubbles, a change in

color, or production of a solid

32


Learning Check

Which of the following represent a chemical change?

A. burning a candle

B. ice melting

C. toast burns

D. water turns to steam

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Solution

Which of the following represent a chemical change?

A. burning a candle chemical change

B. ice melting (this is a physical change)

C. toast burns chemical change

D. water turns to steam (this is a physical change)

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Writing a Chemical Equation

35

To write a chemical equation • an arrow indicates reactants form products

• reactants are written on the left side of the arrow

• products are written on the right side of the arrow

• multiple reactants or products are separated by a + sign

• the delta {Δ} sign indicates that heat is used to start the reaction

• physical states are represented in parenthesis as (s), (l), (g) or (aq) aqueous

Example:

2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g)


Identifying a Balanced Equation

In a balanced chemical equation, • no atoms are lost or gained

• the number of reacting atoms is equal to the number of product atoms

36


Balancing a Chemical Equation

Example: Formation of Al2S3 37

Step 1 Write an equation using the correct formulas of the reactants and products.

Al(s) + S(s) Al2S3(s)

Step 2 Count the atoms of each element in the reactants and products.

Reactants Products

1 atom Al 2 atoms Al Not balanced

1 atoms S 3 atoms S Not balanced

Step 3 Use coefficients to balance each element. Starting with the most

complex formula, change coefficients to balance equation.

2Al(s) + 3S(s) Al2S3(s)

Step 4 Check the final equation to confirm it is balanced.

Make sure coefficients are the lowest ratio.

Reactants Products

2 atom Al 2 atoms Al Balanced !!!

3 atoms S 3 atoms S Balanced !!!


Learning Check

State the number of atoms of each element on the reactant side and the

product side and use coefficients to balance each of the equations:

A. P4(s) + Br2(l) PBr3(g)

B. Al(s) + Fe2O3(s) Fe(s) + Al2O3(s)

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Solution



A. P4(s) + Br2(l) PBr3(g)

B. Al(s) + Fe2O3(s) Fe(s) + Al2O3(s)

39

Before Reactants Products

P atoms 4 1

Br atoms 2 3


Al atoms 1 2

Fe atoms 2 1

O atoms 3 3


Solution



A. P4(s) + 6Br2(l) 4PBr3(g)

B. 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s)

40

After Reactants Products

P atoms 4 4

Br atoms 12 12

After Reactants Products

Al atoms 2 2

Fe atoms 2 2

O atoms 3 3


P atoms 4 1

Br atoms 2 3


Al atoms 1 2

Fe atoms 2 1

O atoms 3 3


Learning Check

Use coefficients to balance each element.

Fe2O3(s) + H2(g) Fe(s) + H2O(l)

41


Solution

Use Coefficients to balance each element.

Fe2O3(s) + 3H2(g) 2Fe(s) + 3H2O(l)

Always check the final equation to confirm it is balanced:

42

Reactants Products

Atoms Fe 2 2

Atoms O 3 3

Atoms H 6 6


Balancing Equations with Polyatomic Ions

Example:

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

43



When balancing equations with polyatomic ions, count each polyatomic ion as a unit.

Example:

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

44

before balance Reactants Products

PO43− ions 1 2

Na+ ions 3 1

Mg2+ ions 1 3

Cl− ions 2 1



When balancing equations with polyatomic ions, balance each polyatomic ion as a unit.

Example:

2Na3PO4(aq) + 3MgCl2(aq) Mg3(PO4)2(s) + 6NaCl(aq)

45

After balance Reactants Products

PO43− ions 2 2

Na+ ions 6 6

Mg2+ ions 3 3

Cl− ions 6 6


Learning Check

Balance and list the coefficients from reactants to products.

__Fe2O3(s) + __C(s) __Fe(s) + __CO2(g)

a) 2, 3, 2, 3 b) 2, 3, 4, 3 c) 1, 1, 2, 3

__Al(s) + __FeO(s) __Fe(s) + __Al2O3(s)

a) 2, 3, 3, 1 b) 2, 1, 1, 1 c) 3, 3, 3, 1

__Al(s) + __H2SO4(aq) __Al2(SO4)3(aq) + __H2(g)

a) 3, 2, 1, 2 b) 2, 3, 1, 3 c) 2, 3, 2, 3

46


Solution

Balance and list the coefficients from reactants to products.

2Fe2O3(s) + 3C(s) 4Fe(s) + 3CO2(g)

Correct answer b) 2, 3, 4, 3

2Al(s) + 3FeO(s) 3Fe(s) + Al2O3(s)

Correct answer a) 2, 3, 3, 1

2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g)

Correct answer b) 2, 3, 1, 3

47


Types of Reactions

Chemical reactions can be classified as

• combination reactions

• decomposition reactions

• single replacement reactions

• double replacement reactions

• neutralization reactions

• combustion reactions

48


Combination Reactions

In a combination reaction,

• two or more elements form one product

• or simple compounds combine to form one product

2Mg(s) + O2(g) 2MgO(s)

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

SO3(g) + H2O(l) H2SO4(aq)

49


Decomposition Reaction

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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Δ

Δ

Most decomposition reactions are endothermic: heat is needed for the reaction to proceed


Single Replacement Reaction

In a single replacement reaction, one element takes the

place of a different element in another reacting a compound.

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)

51


Double Replacement Reaction

In a double replacement, two elements in the reactants

exchange places

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

ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g)

52


Neutralization Reactions

In a neutralization reaction:

An acid react with a base forming a salt and water

base + acid salt + water

Examples of neutralization reactions in “antacids”:

Mg(OH)2(aq) + 2HCl(aq) MgCl2(aq) + 2H2O(l)

Al(OH)3(aq) + 3HCl(aq) AlCl3(aq) + 3H2O(l)

CaCO3(aq) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

NaHCO3(aq) + 2HCl(aq) NaCl(aq) + H2O(l) + CO2(g)

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Milk of Magnesia

Maalox

Pepto-Bismol

Alka-Seltzer


Combustion Reaction

In a combustion reaction,

• a carbon-containing compound burns in oxygen gas to form

carbon dioxide (CO2) and water (H2O)

• energy is released as a product in the form of heat

carbon compound + Oxygen CO2(g) + 2H2O(g) + energy

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + energy

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) + energy

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Combustion reactions are exothermic: heat is generated or released during the chemical reaction


Learning Check

Classify each of the following reactions as combination, decomposition, single replacement, double replacement, or combustion.

2Al(s) + 3H2SO4(aq) Al2(SO4)3(s) + 3H2(g)

Na2SO4(aq) + 2AgNO3(aq) Ag2SO4(s) + 2NaNO3(aq)

4Fe(s) + 3O2(g) 2Fe2O3(s)

C2H4(g) + 2O2(g) 2CO2(g) + 2H2O(g)


55


Solution

Classify each of the following reactions as combination, decomposition, single replacement, double replacement, or combustion.

2Al(s) + 3H2SO4(aq) Al2(SO4)3(s) + 3H2(g)

Na2SO4(aq) + 2AgNO3(aq) Ag2SO4(s) + 2NaNO3(aq)

4Fe(s) + 3O2(g) 2Fe2O3(s)

C2H4(g) + 2O2(g) 2CO2(g) + 2H2O(g)


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Single Replacement

Double Replacement

Combination

Combustion

Neutralization


Learning Check

Identify each reaction as combination, decomposition,

combustion, single replacement, or double replacement.

A. 3Ba(s) + N2(g) Ba3N2(s)

B. 2Ag(s) + H2S(aq) Ag2S(s) + H2(g)

C. 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) + Heat

D. PbCl2(aq) + K2SO4(aq) 2KCl(aq) + PbSO4(s)

E. K2CO3(s) K2O(aq) + CO2(g)

F. Al(OH)3(aq) + 3HCl(aq) AlCl3(aq) + 3H2O(l)

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Δ


Solution

Identify each reaction as combination, decomposition,

combustion, single replacement, or double replacement.

A. 3Ba(s) + N2(g) Ba3N2(s)

B. 2Ag(s) + H2S(aq) Ag2S(s) + H2(g)

C. 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) + Heat

D. PbCl2(aq) + K2SO4(aq) 2KCl(aq) + PbSO4(s)

E. K2CO3(s) K2O(aq) + CO2(g)

F. Al(OH)3(aq) + 3HCl(aq) AlCl3(aq) + 3H2O(l)

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Δ

Combination

Single Replacement

Combustion

Double Replacement

Decomposition

Neutralization


Oxidation−Reduction Reactions

Oxidation–reduction reaction

• provides us with energy from food

• provides electrical energy in batteries

• occurs when iron rusts:

4Fe(s) + 3O2(g) 2Fe2O3(s)

59


Oxidation−Reduction

In an oxidation–reduction reaction, electrons are transferred

from one substance to another.

60


Oxidation and Reduction

2Cu(s) + O2(g) 2CuO(s)

Oxidation is loss of electrons.

2Cu(s) 2Cu2+(s) + 4e−

Reduction is gain of electrons.

O2(g) + 4e− 2O2−(s)

61

The green patina on copper is

due to oxidation


Zn replaces Cu2+

62

CuSO4(aq) + Zn(s) ZnSO4(aq) + Cu(s)

2+ 0 2+ 0


Zn Transfers Electrons to Cu2+

OIL Oxidation is losing electrons.

RIG Reduction is gaining electrons.

63


Learning Check

Identify each of the following as oxidation or reduction:

Sn(s) Sn4+(aq) + 4e−

Fe3+(aq) + 1e− Fe2+

(aq)

Cl2(g) + 2e− 2Cl−(aq)

64


Solution

Identify each of the following as oxidation or reduction:

Sn(s) Sn4+(aq) + 4e−

Fe3+(aq) + 1e− Fe2+

(aq)

Cl2(g) + 2e− 2Cl−(aq)

65

Oxidation

Reduction

Reduction


Learning Check

In light-sensitive sunglasses, UV light initiates an

oxidation–reduction reaction.

UV light

2Ag+ + 2Cl− 2Ag + Cl2

A. Which reactant is oxidized?

B. Which reactant is reduced?

66


Solution

In light-sensitive sunglasses, UV light initiates an

oxidation-reduction reaction.

uv light

2Ag+ + 2Cl− 2Ag + Cl2

A. Which reactant is oxidized? Chlorine

2Cl− Cl2 + 2e−

B. Which reactant is reduced? Silver

2Ag+ + 2e− 2Ag

67


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

• the mass of products is equal to mass of reactants

68


Conservation of Mass

69

2(107.9 g) + 32.07 g = 247.87 g

Matter cannot be created or

destroyed, therefore the mass of

products is equal to mass of reactants

2 moles Ag + 1 mole S = 1 mole Ag2S


Reading Equations in Moles

Consider the following equation:

2Fe(s) + 3S(s) Fe2S3(s)

An equation can be read in “moles” by placing the word “moles of” between each coefficient and formula.

2 moles of Fe + 3 moles of S 1 mole of Fe2S3

70

2 (55.85 g) + 3 (32.07 g) 207.91 g


Writing Conversion Factors from Equations

A mole–mole factor is a ratio of the moles for any two substances in an equation.

2Fe(s) + 3S(s) Fe2S3(s)

Fe and S 2 moles Fe or 3 moles S

3 moles S 2 moles Fe

Fe and Fe2S3 2 moles Fe or 1 moles Fe2S3

1 mole Fe2S3 2 moles Fe

S and Fe2S3 3 moles S or 1 mole Fe2S3

1 mole Fe2S3 3 moles S

71


Calculations with Mole Factors

Example:

How many moles of Fe2O3 can form from 6.0 moles of O2?

4Fe(s) + 3O2(g) 2Fe2O3(s)

72


Calculations with Mole Factors

How many moles of Fe2O3 can form from 6.0 moles of O2?

4Fe(s) + 3O2(g) 2Fe2O3(s)

Relationship: 3 moles of O2 = 2 moles of Fe2O3

Use a mole–mole factor to determine the moles of Fe2O3.

moles of Fe2O3 = 6.0 moles O2 x 2 moles Fe2O3

3 moles O2

= 4.0 moles of Fe2O3

73


Learning Check

How many moles of Fe are needed for the reaction of

12.0 moles of O2?

4Fe(s) + 3O2(g) 2Fe2O3(s)

A. 3.00 moles of Fe

B. 9.00 moles of Fe

C. 16.0 moles of Fe

74


Solution

How many moles of Fe are needed for the reaction of 12.0 moles of O2?

4Fe(s) + 3O2(g) 2Fe2O3(s)

The possible mole factors for the solution are:

4 moles Fe or 3 moles O2

3 moles O2 4 moles Fe

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Set up the problem using the mole–mole factor that cancels given moles.

moles Fe = 12.0 moles O2 x 4 moles Fe = 16.0 moles Fe

3 moles O2

The answer is C, 16.0 moles Fe


Converting from Moles to Grams

76

Determine the mass (g) of NH3 that can be produced from 32 grams of N2. N2(g) + 3H2(g) 2NH3(g)

Hint: In this case, question and data refers to grams (instead of moles)

therefore the moles need to be converted to grams !!!!


Converting from Moles to Grams

77

Determine the mass (g) of NH3 that can be produced from 32 grams of N2. N2(g) + 3H2(g) 2NH3(g)

N2(g) + 3H2(g) 2NH3(g)

28.02g 3(2.016g) 2(17.034g)

Converting to grams:

1 mole N2(g) + 3 moles H2(g) 2 moles NH3(g)

28.02g N2(g) + 6.048g H2(g) 34.068g NH3(g)

Therefore:

grams of NH3 = 32 grams of N2 x 34.068 grams of NH3

28.02 grams of N2

grams of NH3 = 38.9071 g = 39 g of NH3


Learning Check

How many grams of O2 are needed to produce 45.8 grams of

Fe2O3 in the following reaction?

4Fe(s) + 3O2(g) 2Fe2O3(s)

A. 38.4 g of O2

B. 13.8 g of O2

C. 1.38 g of O2

78


Solution

Step 1 Use molar mass to convert moles to grams:

79



4Fe(s) + 3O2(g) 2Fe2O3(s)

4Fe(s) + 3O2(g) 2Fe2O3(s)

4 moles Fe(s) + 3 moles O2(g) 2 moles Fe2O3(s)

4(55.85g) 3(32.00g) 2(159.7g)

223.4g 96.00g 319.4g

Now you can resolve the problem using a gram-to-gram

relationship for the above given chemical equation


Solution

80



4Fe(s) + 3O2(g) 2Fe2O3(s)

4 moles Fe(s) + 3 moles O2(g) 2 moles Fe2O3(s)

Therefore: grams of O2 = 45.8 grams Fe2O3 x 96.00 grams O2

319.4 grams Fe2O3

grams of O2 = 13.7658g

grams of O2 = 13.8g O2 (the correct answer is B)

4(55.85g) 3(32.00g) 2(159.7g)

223.4g 96.00g 319.4g


Energy in Chemical Reactions

Exothermic Reactions

In an exothermic reaction, • heat is released

• the energy of the products is less

than the energy of the reactants

• heat is a product

C(s) + 2H2(g) CH4(g) + 18 kcal

81


In an endothermic reaction,

• heat is absorbed

• the energy of the products is greater

than the energy of the reactants

• heat is a reactant (added)

82

Energy in Chemical Reactions

Endothermic Reactions

N2(g) + O2(g) + 43.3 kcal 2NO(g)


Learning Check

Identify each reaction as exothermic or endothermic.

A. N2(g) + 3H2(g) 2NH3(g) + 22 kcal

B. CaCO3(s) + 133 kcal CaO(s) + CO2(g)

C. 2SO2(g) + O2(g) 2SO3(g) + heat

83


Solution

Identify each reaction as exothermic or endothermic.

A. N2(g) + 3H2(g) 2NH3(g) + 22 kcal Exothermic

B. CaCO3(s) + 133 kcal CaO(s) + CO2(g)

Endothermic

C. 2SO2(g) + O2(g) 2SO3(g) + heat

Exothermic

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Reaction Rate

The reaction rate is the speed at which reactant is used up

The reaction rate is the speed at which product forms

The reaction rate

• increases when temperature rises because reacting molecules move

faster, providing more colliding molecules with energy of activation

• increases when agitation is used (if reactants are solids or liquids)

because reacting molecules move faster (more collisions)

• increases with increase in concentration of reactants

• increases when a catalyst is used (catalysts are chemical compounds

that lowers the energy of activation, therefore facilitating the

formation of product molecules (reaction products are formed faster)

85


Catalyst

A catalyst

• increases the rate of a reaction

• lowers the energy of activation

• is not used up during the reaction

86


Factors That Increase Reaction Rate

87

Use agitation More collisions

among molecules


Learning Check

State the effect of each on the rate of reaction as increases,

decreases, or has no effect:

A. increasing the temperature

B. removing some of the reactants

C. adding a catalyst

D. placing the reaction flask in ice

E. increasing the concentration of one of the reactants

F. using a glass rod to mix the liquid reactants

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Solution

State the effect of each on the rate of reaction as

increases, decreases, or has no effect:

A. increasing the temperature increases

B. removing some of the reactants decreases

C. adding a catalyst increases

D. placing the reaction flask in ice decreases

E. increasing the concentration of one of the reactant increases

F. using a glass rod to mix the liquid reactants increases

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Learning Check

Indicate the effect of each factor listed on the rate of the

following reaction as increases, decreases, or no change.

2CO(g) + O2(g) 2CO2(g)

A. raising the temperature

B. adding O2

C. adding a catalyst

D. lowering the temperature

90


Solution

Indicate the effect of each factor listed on the rate of the

following reaction as increases, decreases, or no change.

2CO(g) + O2(g) 2CO2(g)

A. raising the temperature increases

B. adding O2 increases

C. adding a catalyst increases

D. lowering the temperature decreases

91

chapter 5 chemical quantities and reactions

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