1

Chapter 5 Chemical Reactions

5.9a The Mole Relationships in

Chemical Equations

2

Law of Conservation of Mass and Stoichiometry

The Law of Conservation of Mass indicates that in an

ordinary chemical reaction, • Matter cannot be created nor destroyed.• No change in total mass occurs in a reaction.

• Mass of products is equal to mass of reactants.

Stoichiometry is the study of mass relationships in chemical reactions.

3

Conservation of Mass

2 moles Ag + 1 moles S = 1 mole Ag2S

2 (107.9 g) + 1(32.1 g) = 1 (247.9 g)

247.9 g reactants = 247.9 g product

4

Consider the following equation:4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

This equation can be read in “moles” by placing theword “moles” between each coefficient and formula.

4 moles Fe + 3 moles O2 2 moles Fe2O3

Reading Equations In Moles

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A mole-mole factor is a ratio of the moles for any two

substances in an equation.

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

Fe and O2 4 moles Fe and 3 moles O2

3 moles O2 4 moles Fe

Fe and Fe2O3 4 moles Fe and 2 moles Fe2O3

2 moles Fe2O3 4 moles Fe

O2 and Fe2O3 3 moles O2 and 2 moles Fe2O3

2 moles Fe2O3 3 moles O2

Writing Mole-Mole Factors

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Consider the following equation:

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

A. A mole-mole factor for H2 and N2 is

1) 3 moles N2 2) 1 mole N2 3) 1 mole N2

1 mole H2 3 moles H2 2 moles H2

B. A mole-mole factor for NH3 and H2 is

1) 1 mole H2 2) 2 moles NH3 3) 3 moles N2

2 moles NH3 3 moles H2 2 moles NH3

Learning Check

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

A. A mole-mole factor for H2 and N2 is 2) 1 mole N2

3 moles H2

B. A mole-mole factor for NH3 and H2 is

2) 2 moles NH3

3 moles H2

Solution

8

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

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

Relationship: 3 mole O2 = 2 mole Fe2O3

Write a mole-mole factor to determine the moles of Fe2O3.

6.0 mole O2 x 2 mole Fe2O3 = 4.0 mole Fe2O3

3 mole O2

Calculations with Mole Factors

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Guide to Using Mole Factors

10

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

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

1) 3.00 moles Fe 2) 9.00 moles Fe3) 16.0 moles Fe

Learning Check

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3) 16.0 moles Fe

12.0 moles O2 x 4 moles Fe = 16.0 moles Fe 3 moles O2

Solution