Chapter 8 Quantities in Chemical Reactions

2006, Prentice Hall

Octane in gas tank

Octane mixeswith oxygen

Products arecarbon dioxideand water

2

CHAPTER OUTLINE

Stoichiometry Molar Ratios Mole-Mole Calculations Mass-Mole Calculations Mass-Mass Calculations Limiting Reactant Percent Yield

Global Warming: Too Much Carbon Dioxide

• The combustion of fossil fuels such as octane (shown here) produces water and carbon dioxide as products.

• Carbon dioxide is a greenhouse gas that is believed to be responsible for global warming.

  The greenhouse effect

• Greenhouse gases act like glass in a greenhouse, allowing visible-light energy to enter the atmosphere but preventing heat energy from escaping.

• Outgoing heat is trapped by greenhouse gases.

Combustion of fossil fuels produces CO2.

• Consider the combustion of octane (C8H18), a component of gasoline:

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)

• The balanced chemical equation shows that 16 mol of CO2 are produced for every 2 mol of octane burned.

Global Warming

• scientists have measured an average 0.6°C rise in atmospheric temperature since 1860

• during the same period atmospheric CO2 levels have risen 25%

The Source of Increased CO2

• the primary source of the increased CO2 levels are combustion reactions of fossil fuels we use to get energy (methane and octane)– 1860 corresponds to the beginning of the

Industrial Revolution in the US and Europe

(g)(g)(g)(g) OH 2 CO O 2 CH 2224 (g)(g)(g)(l) OH 18 CO 16 O 25 HC 2 222188

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STOICHIOMETRY

Stoichiometry is the quantitative relationship between the reactants and products in a balanced chemical equation.

A balanced chemical equation provides several important information about the reactants and products in a chemical reaction.

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MOLARRATIOS

For example:

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

1 molecule 3 molecules 2 molecules

100 molecules 300 molecules 200 molecules

106 molecules 3x106 molecules 2x106 molecules

1 mole 3 moles 2 moles

This is the molar ratios between

the reactants and products

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Examples:

2

2

mol O=

mol CO 13

8

4 10

2

mol C H=

mol H O

Determine each mole ratio below based on the reaction shown:

2 C4H10 + 13 O2 8 CO2 + 10 H2O

2

10

Stoichiometry - Allows us to predict products that form in a reaction based on amount of reactants.

Stoichiometry

The amount of each element must be the same throughout the overall reaction.

For example, the amount of element H or O on the reactant side mustequal the amount of element H or O on the product side.

2H2 + O2 2H2O

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STOICHIOMETRICCALCULATIONS

Stoichiometric calculations can be classified as one of the following:

MOLES of compound A

MOLES of compound B

molar ratio

Mole-mole calculations

MASS of compound A

MM

Mass-mole calculations

MASS of compound B

MM

Mass-mass calculations

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MOLE-MOLECALCULATIONS

Relates moles of reactants and products in a balanced chemical equation

MOLES of compound A

MOLES of compound B

molar ratio

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How many moles of ammonia can be produced from 32 moles of hydrogen? (Assume excess N2 present)

Example 1:

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

32 mol H23

2

mol NHx

mol H23 = 21 mol NH3

Mole ratio

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In one experiment, 6.80 mol of ammonia are prepared. How many moles of hydrogen were used up in this experiment?

Example 2:

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

6.80 mol NH32

3

mol Hx

mol NH32 = 10.2 mol H2

Mole ratio

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MASS-MOLECALCULATIONS

Relates moles and mass of reactants or products in a balanced chemical equation

MOLES of compound A

MOLES of compound B

molar ratio

MASS of compound A

MM

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How many moles of ammonia can be produced from the reaction of 125 g of nitrogen?

Example 1:

2

2

mol Nx

g N

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

125 g N2

1

28.0= 8.93 mol NH3

3

2

mol NHx

mol N12

Molar mass Mole ratio

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MASS -MASSCALCULATIONS

Relates mass of reactants and products in a balanced chemical equation

MOLES of compound A

MOLES of compound B

molar ratio

MASS of compound A

MM

MASS of compound B

MM

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What mass of carbon dioxide will be produced from the reaction of 175 g of propane, as shown?

Example 1:

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

Mass of propane

Moles of propane

Moles of carbon dioxide

Mass of carbon dioxide

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Example 1:

3 8

3 8

mol C Hx

g C H2

3 8

mol COx

mol C H

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

175 g C3H8 44.1

1

= 524 g CO2

1

3

2

2

g COx

mol CO

Molar massMole ratio

44.01

Molar mass

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LIMITINGREACTANT

When 2 or more reactants are combined in non-stoichiometric ratios, the amount of product produced is limited by the reactant that is not in excess.

This reactant is referred to as limiting reactant. When doing stoichiometric problems of this

type, the limiting reactant must be determined first before proceeding with the calculations.

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LIMITING REACTANT ANALOGY

Consider the following recipe for a sundae:

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LIMITING REACTANT ANALOGY

How many sundaes can be prepared from the following ingredients:The number of sundaes possible is limited by the amount of syrup, the limiting reactant.

Limiting reactant

Excess reactants

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LIMITINGREACTANT

When solving limiting reactant problems, assume each reactant is limiting reactant, and calculate the desired quantity based on that assumption.

A + B C

A is LR

B is LR

Calculate amount of C

Calculate amount of C

Compare your answers for each assumption; the lower value is the correct assumption.Lower

value is correct

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A fuel mixture used in the early days of rocketry was a mixture of N2H4 and N2O4, as shown below. How many grams of N2 gas is produced when 100 g of N2H4 and 200 g of N2O4 are mixed?

Example 1:

2 N2H4 (l) + 1 N2O4 (l) 3 N2 (g) + 4 H2O (g)

Limiting reactant

Mass-mass calculations

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Example 1:

2 4

2 4

mol N Hx

g N H2

2 4

mol Nx =

mol N H100 g N2H4

32.04

4.68 mol N2

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2 N2H4 (l) + 1 N2O4 (l) 3 N2 (g) + 4 H2O (g)

Assume N2H4 is LR

1

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Example 1:

2 4

2 4

mol N Ox

g N O2

2 4

mol Nx =

mol N O200 g N2O4

92.00

6.52 mol N2

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2 N2H4 (l) + 1 N2O4 (l) 3 N2 (g) + 4 H2O (g)

Assume N2O4 is LR

1

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Example 1:

6.52 mol N2

2 N2H4 (l) + 1 N2O4 (l) 3 N2 (g) + 4 H2O (g)

Assume N2O4 is LR

Assume N2H4 is LR 4.68 mol N2

Correct amount

N2H4 is LR

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Example 1:

2

2

g Nx

mol N4.68 mol N2

28.0= 131 g N2

2 N2H4 (l) + 1 N2O4 (l) 3 N2 (g) + 4 H2O (g)

Calculate mass of N2

1

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How many grams of AgBr can be produced when 50.0 g of MgBr2 is mixed with 100.0 g of AgNO3, as shown below:

Example 2:

MgBr2 + 2 AgNO3 2 AgBr + Mg(NO3)2

Limiting Reactant

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Example 2:

2

2

mol MgBrx

g MgBr 2

mol AgBrx

mol MgBr50.0 g MgBr2 184.1 1

2

Assume MgBr2 is LR

1

MgBr2 + 2 AgNO3 2 AgBr + Mg(NO3)2

g AgBrx =

mol AgBr187.8

1 102 g AgBr

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Example 2:

3

3

mol AgNOx

g AgNO 3

mol AgBrx

mol AgNO100.0 g AgNO3 169.9 22

Assume AgNO3 is LR

1

MgBr2 + 2 AgNO3 2 AgBr + Mg(NO3)2

g AgBrx =

mol AgBr187.8

1 111 g AgBr

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Example 2:

111 g AgBrAssume AgNO3 is LR

Assume MgBr2 is LR 102 g AgBr

Correct amount

MgBr2 is LR

MgBr2 + 2 AgNO3 2 AgBr + Mg(NO3)2

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PERCENT YIELD

The amount of product calculated through stoichiometric ratios are the maximum amount product that can be produced during the reaction, and is thus called theoretical yield.

The actual yield of a product in a chemical reaction is the actual amount obtained from the reaction.

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PERCENT YIELD

The percent yield of a reaction is obtained as follows:

Actual yieldx100 = Percent yield

Theoretical yield

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In an experiment forming ethanol, the theoretical yield is 50.0 g and the actual yield is 46.8 g. What is the percent yield for this reaction?

Example 1:

Actual yield% yield = x100

Theoretical yield

46.8 g= x100 =

50.0 g 92.7 %

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Silicon carbide can be formed from the reaction of sand (SiO2) with carbon as shown below:

Example 2:

1 SiO2 (s) + 3 C (s) 1 SiC (s) + 2 CO (g)

When 100 g of sand are processed, 51.4g of SiC is produced. What is the percent yield of SiC in this reaction?

Actual yield

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Example 2:

2

2

mol SiOx

g SiO 2

mol SiCx

mol SiO100 g SiO2 60.1

66.7 g SiC

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Calculate theoretical yield

1

1 SiO2 (s) + 3 C (s) 1 SiC (s) + 2 CO (g)

g SiCx =

mol SiC140.1

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Example 2:

Actual yield% yield = x100

Theoretical yield

51.4 g= x100 =

66.7 g 77.1 %

Calculate percent yield

Theoretical and Actual Yield• In order to determine the theoretical yield, we

use reaction stoichiometry to determine the amount of product each of our reactants could make.

• The theoretical yield will always be the least possible amount of product.– The theoretical yield will always come from the

limiting reactant.

• Because of both controllable and uncontrollable factors, the actual yield of product will always be less than the theoretical yield.

1. Limiting reactant - the reactant that limits the amount of product produced in a chemical reaction. The reactant that makes the least amount of product.

2. Theoretical yield - the amount of product that can be made in a chemical reaction based on the amount of limiting reactant.

3. Actual yield - the amount of product actually produced by a chemical reaction.

4. Percent yield - The percent of the theoretical yield that was actually obtained.

% yield = actual yield

theoretical yieldx 100

Chap. 8 terms you should know

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THE END