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Chapter 5 Chemical Reactions and Quantities

5.7 Mole Relationships in Chemical Equations

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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 in a reaction. §  Mass of products is equal to mass of reactants.

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Conservation of Mass

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Reactants Products 2 moles Ag + 1 moles S = 1 mole Ag2S 2 (107.9 g) + 1(32.1 g) = 1 (247.9 g) 247.9 = 247.9 g

Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

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

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

4 moles Fe + 3 moles O2 2 moles Fe2O3

Reading Equations with 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

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How many moles of Fe2O3 can be produced from 6.0 moles 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 moles Fe2O3

3 mole O2

Calculations with Mole Factors

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Guide to Using Mole Factors

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

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

1) 3.00 moles Fe 2) 9.00 moles Fe 3) 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

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Chapter 5 Chemical Reactions and Quantities

5.8 Mass Calculations for Reactions

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Moles to Grams

Suppose we want to determine the mass (g) of NH3

that can form from 2.50 moles N2. N2(g) + 3H2(g) 2NH3(g) The plan needed would be moles N2 moles NH3 grams NH3 The factors needed would be: mole factor NH3/N2

and the molar mass NH3

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Moles to Grams

The setup for the solution would be: 2.50 mole N2 x 2 moles NH3 x 17.0 g NH3

1 mole N2 1 mole NH3 given mole-mole factor molar mass

= 85.0 g NH3

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How many grams of O2 are needed to produce 0.400 mole Fe2O3 in the following reaction?

4Fe(s) + 3O2(g) 2Fe2O3(s) 1) 38.4 g O2 2) 19.2 g O2 3) 1.90 g O2

Learning Check

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2) 19.2 g O2 0.400 mole Fe2O3 x 3 mole O2 x 32.0 g O2= 19.2 g O2

2 mole Fe2O3 1 mole O2

mole factor molar mass

Solution

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The reaction between H2 and O2 produces 13.1 g water. How many grams of O2 reacted?

2H2(g) + O2(g) 2H2O(g) ? g 13.1 g

The plan and factors would be g H2O mole H2O mole O2 g O2 molar mole-mole molar mass H2O factor mass O2

Calculating the Mass of a Reactant

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The setup would be: 13.1 g H2O x 1 mole H2O x 1 mole O2 x 32.0 g O2

18.0 g H2O 2 moles H2O 1 mole O2 molar mole-mole molar

mass H2O factor mass O2

= 11.6 g O2

Calculating the Mass of a Reactant

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Learning Check

Acetylene gas C2H2 burns in the oxyacetylene torch for welding. How many grams of C2H2 are burned if the reaction produces 75.0 g CO2?

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)

1) 88.6 g C2H2 2) 44.3 g C2H2 3) 22.2 g C2H2

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3) 22.2 g C2H2

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g)

75.0 g CO2 x 1 mole CO2 x 2 moles C2H2 x 26.0 g C2H2 44.0 g CO2 4 moles CO2 1 mole C2H2

molar mole-mole molar mass CO2 factor mass C2H2

= 22.2 g C2H2

Solution

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Calculating the Mass of Product

When 18.6 g ethane gas C2H6 burns in oxygen, how many grams of CO2 are produced?

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) 18.6 g ? g The plan and factors would be g C2H6 mole C2H6 mole CO2 g CO2 molar mole-mole molar mass C2H6 factor mass CO2

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Calculating the Mass of Product

The setup would be

18.6 g C2H6 x 1 mole C2H6 x 4 moles CO2 x 44.0 g CO2 30.1 g C2H6 2 moles C2H6 1 mole

CO2

molar mole-mole molar mass C2H6 factor mass CO2

= 54.4 g CO2

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Learning Check

How many grams H2O are produced when 35.8 g C3H8

react by the following equation? C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

1. 14.6 g H2O 2. 58.4 g H2O 3. 117 g H2O

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Solution

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 35.8 g C3H8x 1 mole C3H8 x 4 mole H2O x 18.0 g H2O 44.1 g C3H8 1 mole C3H8 1 mole H2O

molar mole-mole molar mass C3H8 factor mass H2O = 58.4 g H2O (2)

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Chapter 5 Chemical Reactions and Quantities

5.9 Percent Yield and Limiting Reactants

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Theoretical, Actual, and Percent Yield

Theoretical yield §  The maximum amount of product, which is calculated

using the balanced equation. Actual yield §  The amount of product obtained when the reaction takes

place. Percent yield §  The ratio of actual yield to theoretical yield.

percent yield = actual yield (g) x 100 theoretical yield (g)

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Guide to Calculations for Percent Yield

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You prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While you talk, a sheet of 12 cookies burn and you throw them out. The rest of the cookies are okay. What is the percent yield of edible cookies?

Theoretical yield 60 cookies possible Actual yield 48 cookies to eat Percent yield 48 cookies x 100 = 80% yield

60 cookies

Calculating Percent Yield

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Learning Check

Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide. 2C(g) + O2(g) 2CO(g)

What is the percent yield if 40.0 g CO are produced when 30.0 g O2 are used?

1) 25.0%

2) 75.0%

3) 76.2% 30

Solution

3) 76.2 % yield theoretical yield of CO 30.0 g O2 x 1 mole O2 x 2 moles CO x 28.0 g CO 32.0 g O2 1 mole O2 1 mole CO

= 52.5 g CO (theoretical) percent yield

40.0 g CO (actual) x 100 = 76.2 % yield 52.5 g CO (theoretical)

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Learning Check

When N2 and 5.00 g H2 are mixed, the reaction produces 16.0 g NH3. What is the percent yield for

the reaction?

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

1) 31.3 % 2) 56.5 % 3) 80.0 %

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Solution

2) 56.5 % N2(g) + 3H2(g) 2NH3(g) 5.00 g H2 x 1 mole H2 x 2 moles NH3 x 17.0 g NH3 2.02 g H2 3 moles H2 1 mole NH3

= 28.2 g NH3 (theoretical) Percent yield = 16.0 g NH3 x 100 = 56.7 %

28.2 g NH3

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Limiting Reactant

A limiting reactant in a chemical reaction is the substance that

§  Is used up first.

§  Limits the amount of product that can form and stops the reaction.

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Reacting Amounts

In a table setting, there is 1plate, 1 fork, 1 knife, and 1 spoon. How many table settings are possible from 5 plates, 6 forks, 4 spoons, and 7 knives? What is the limiting item?

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Reacting Amounts

Only 4 place settings are possible.

Initially Used Left over Plates 5 4 1 Forks 6 4 2 Spoons 4 4 0 Knives 7 4 3 The limiting item is the spoon.

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Example of An Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices of bread and 1 jar of peanut butter?

With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item.

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Example of An Everyday Limiting Reactant

How many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter? With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item.

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Guide to Calculating Product from a Limiting Reactant

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Limiting Reactant

When 4.00 moles H2 is mixed with 2.00 moles Cl2,how many moles of HCl can form? H2(g) + Cl(g) ⎯→ 2HCl (g)

4.00 moles 2.00 moles ??? moles

§  Calculate the moles of product that each reactant, H2 and Cl2, could produce.

§  The limiting reactant is the one that produces the smallest amount of product.

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Limiting Reactant

HCl from H2 4.00 moles H2 x 2 moles HCl = 8.00 moles HCl

1 moles H2 (not possible) HCl from Cl2 2.00 moles Cl2 x 2 moles HCl = 4.00 moles HCl

1 mole Cl2 (smaller number of moles, Cl2 will be used up first)

The limiting reactant is Cl2 because it is used up first. Thus Cl2 produces the smaller number of moles of HCl.

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Check Calculations

Initially H2

4.00 moles

Cl2

2.00 moles

2HCl 0 mole

Reacted/ Formed

-2.00 moles -2.00 moles +4.00 moles

Left after reaction

2.00 moles Excess

0 mole Limiting

4.00 moles

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Limiting Reactants Using Mass

If 4.80 moles Ca are mixed with 2.00 moles N2, which is The limiting reactant? 3Ca(s) + N2(g) ⎯→ Ca3N2(s)

moles of Ca3N2 from Ca 4.80 moles Ca x 1 mole Ca3N2 = 1.60 moles Ca3N2

3 moles Ca (Ca is used up) moles of Ca3N2 from N2 2.00 moles N2 x 1 mole Ca3N2 = 2.00 moles Ca3N2

1 mole N2 (not possible) Ca is used up when 1.60 mole Ca3N2 forms. Thus, Ca is the limiting reactant.

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Limiting Reactants Using Mass

Calculate the mass of water produced when 8.00 g H2

and 24.0 g O2 react?

2H2(g) + O2(g) 2H2O(l)

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Limiting Reactants Using Mass

Moles H2O from H2: 8.00 g H2 x 1 mole H2 x 2 moles H2O = 3.97 moles H2O

2.02 g H2 2 moles H2 (not possible)

Moles H2O from O2:

24.0 g O2 x 1 mole O2 x 2 moles H2O = 1.50 moles H2O 32.0 g O2 1 mole O2 O2 is limiting The maximum amount of product is 1.50 moles H2O, which is converted to grams. 1.50 moles H2O x 18.0 g H2O = 27.0 g H2O 1 mole H2O