Topic 9: Stoichiometry

[Pages:13]Topic 9: Stoichiometry

(Chapter 9 in Modern Chemistry p. 298)

Introduction to Stoichiometry

In this topic we will focus on the quantitative aspects of chemical reactions. That's right; you'll need your calculators. Composition Stoichiometry (Topic 7) deals with the mass relationships of elements in compounds. Reaction Stoichiometry involves the mass relationships between reactants and products in a chemical reaction. Reaction stoichiometry is the subject of this topic and it is based on chemical equations and the law of conservation of mass. All reaction stoichiometry calculations must start with a balanced chemical equation.

Reaction Stoichiometry Problems

Problems in reaction stoichiometry will be classified by the given and the unknown.

Problem Type 1: Given & unknown quantities are amounts in moles.

amount of given substance (mol)

amount of unknown substance (mol)

Problem Type 2: Given amount is in moles and unknown amount is mass in grams.

amount of given substance (mol)

amount of unknown substance (mol)

mass of unknown substance (g)

Problem Type 3: Given amount is mass in grams and unknown amount is in moles.

mass of given substance (g)

amount of given substance (mol)

amount of unknown substance (mol)

Problem Type 4: Given & unknown amounts are mass in grams.

mass of given substance (g)

amount of given substance (mol)

amount of unknown substance (mol)

mass of unknown substance (g)

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Mole Ratio

Solving any reaction stoichiometry problem requires the use of a mole ratio to convert from mores or grams of one substance in a reaction to moles or grams of another substance. A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction. This information is obtained directly from the balanced chemical equation.

For example, here is the equation for the electrolysis of melted aluminum oxide to produce aluminum and oxygen.

2 Al2O3(l) 4 Al(s) + 3 O2(g)

This means: 2 moles of Al2O3 can produce 4 moles of Al and 3 moles of O2

This can be represented as a mole ratio like:

2 mol Al2O3

or

4 mol Al

4 mol Al

2 mol Al2O3

2 mol Al2O3

or

3 mol O2

3 mol O2

2 mol Al2O3

4 mol Al

or

3 mol O2

3 mol O2

4 mol Al

Can you see that this would also mean: 4 moles of Al2O3 can produce 8 moles of Al and 6 moles of O2

Example Problem 1: Determine the amount in moles of aluminum that can be produced from 13.0 mol of aluminum oxide.

What is the mole ratio that you will need? Al to Al2O3

The Al produce is 2x as many moles as aluminum oxide. So the answer is simply 26.0 moles of Al.

It's not always so easy to do the math in your head. If you need to you can use dimensional analysis.

13.0 mol Al2O3

x

4 mol Al

=

2 mol Al2O3

26.0 mol Al

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Note: You will always use a mole ratio when you need to change from one substance in an equation to another substance in that equation. Molar Mass

Remember in Topic 7 that the molar mass is the mass, in grams, of one mole of a substance. Recall that the molar mass can also be used as a conversion factor that relates the mass of a substance to the amount in moles of that same substance. You can use molar mass to go from grams to moles or to go from moles to grams.

Again using the equation on p. 2:

1 mol Al2O3 = 101.96 g 1 mol Al = 26.98 g 1 mol O2 = 32.00 g

These molar masses can be expressed by the following conversion factors.

101.96 g Al2O3

or

1 mol Al2O3

1 mol Al2O3

101.96 g Al2O3

26.98 g Al

or

1 mol Al

1 mol Al

26.98 g Al

32.00 g O2

or

1 mol O2

1 mol O2

32.00 g O2

Example Problem 2: Determine the grams of aluminum equivalent to 26.0 moles of aluminum.

26.0 mol Al x

26.98 g Al

=

1 mol Al

701 g Al

Ideal Stoichiometric Calculations

Chemical equations help us make predictions about chemical reactions without having to run the reactions in the laboratory. These reactions are theoretical. They tell us the amounts of reactants and products for a given chemical reaction under ideal conditions, in which all reactants are completely converted into products. Theoretical stoichiometric calculations allow us to determine the maximum amount of product that could be obtained in a reaction.

These problems are extensions of the composition stoichiometry problems that were solved in Topic 7 (mole conversions). It is important to use a logical, systematic approach to successfully solve these problems. We will use dimensional analysis.

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Conversion of Quantities in Moles (Mole-Mole stoichiometry)

The Plan:

amount of given substance (mol)

amount of unknown substance (mol)

To solve this plan requires one conversion factor ? the mole ratio of the unknown to the given from the balanced equation.

Mole ratio (Balanced equation)

Amount of

given substance

x

(mol)

mol unknown mol given

Amount of

=

unknown substance

(mol)

GIVEN IN THE PROBLEM

CONVERSION FACTOR

CALCULATED

Example Problem 3 (p. 305): In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by its reaction with lithium hydroxide, LiOH, according to the following chemical equation.

CO2(g) + 2 LiOH(s) Li2CO3(s) + H2O(l)

How many moles of lithium hydroxide are required to react with 20 moles of CO2, the average amount exhaled by a person each day?

20 mol CO2 Given

x

2 mol LiOH

=

1 mol CO2

Conversion factor from balanced equation

40 mol LiOH

Calculated answer

Task 9a

1. Ammonia, NH3, is widely used as a fertilizer and in many household cleaners. How many moles of ammonia are produced when 6 mol of hydrogen gas react with an excess of nitrogen gas?

2. The decomposition of potassium chlorate is used as a source of oxygen in the laboratory. How many moles of potassium chlorate are needed to produce 15 mole of oxygen gas?

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Conversions of Amounts in Moles to Mass (Mole-Mass Stoichiometry)

The Plan: amount of

given substance (mol)

amount of unknown substance

(mol)

mass of unknown substance

(g)

To solve this plan two conversion factors are required ? the mole ratio of the unknown to the given from the balanced equation and the molar mass of the unknown.

Mole ratio (Balanced equation)

Molar mass factor (Periodic table)

Amount of

given substance

x

(mol)

mol unknown mol given

x Molar mass of unknown (in g) =

Mass of unknown

1 mol of unknown

substance (g)

GIVEN IN THE

PROBLEM

CONVERSION FACTORS

CALCULATED

Example Problem 4 (p. 306): In photosynthesis, plants use energy from the sun to produce glucose, C6H12O6, and oxygen from the reaction of carbon dioxide and water. What mass, in grams, of glucose is produced when 3.00 mol of water react with carbon dioxide?

First, write a balanced equation 6 CO2(g) + 6 H2O(l) C6H12O6(s) + 6 O2(g)

3.00 mol H2O

x

1 mol C6H12O6 6 mol H2O

x

180.18 g C6H12O6 1 mol C6H12O6

=

90.1 g C6H12O6

Given

Conversion factor from balanced equation

Molar mass from the periodic table

Calculated answer

Task 9b

1. When magnesium burns in air, it combines with oxygen to form magnesium oxide. What mass in grams of magnesium oxide is produced from 2.00 mol of magnesium?

2. What mass of glucose can be produced from a photosynthesis reaction that occurs using 10 mol CO2?

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Conversion of Mass to Amounts in Moles (Mass-Mole Stoichiometry)

The Plan: mass of

given substance (g)

amount of given substance

(mol)

amount of unknown substance

(mol)

To solve this plan two conversion factors are required ? the molar mass of the unknown and the mole ratio.

Molar mass factor (Periodic table)

Mole ratio (Balanced equation)

Mass of given

substance (g)

x

1 mol given Molar mass of given (in g)

x

mol unknown mol given

Amount of

=

unknown substance

(mol)

GIVEN IN THE

PROBLEM

CONVERSION FACTORS

CALCULATED

Example Problem 5 (p. 309): The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia.

NH3(g) + O2(g) NO(g) + H2O(g) The reaction is run using 824 g ammonia and excess oxygen. How many moles of H2O are formed?

Even though we were given the equation, it is not balanced. Balance the equation. 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

824 g NH3 x

1 mol NH3

x

17.04 g NH3

6 mol H2O 4 mol NH3

= 72.5 mol H2O

Given

Molar mass from the periodic table

Conversion factor from balanced equation

Calculated answer

Task 9c

1. How many moles or mercury(II) oxide, HgO, is required to decompose to produce 124 g of oxygen?

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Mass-Mass Calculations

Mass-mass calculations are more practical than other mole calculations. You can never measure moles directly but you can measure mass directly. Mass-mass calculations can be viewed as the combination of the other types of problems.

The Plan: mass of

given substance (g)

amount of given substance

(mol)

amount of unknown substance

(mol)

amount of unknown substance

(g)

To solve this plan two conversion factors are required ? the molar mass of the unknown and the mole ratio.

Molar mass factor (Periodic table)

Mole ratio (Balanced equation)

Molar mass factor (Periodic table)

Mass of given

substance (g)

x

1 mol given Molar mass of given

x

(in g)

mol unknown mol given

Molar mass of

x

unknown (in g)

Amount of = unknown

1 mol unknown

substance (mol)

GIVEN IN THE

PROBLEM

CONVERSION FACTORS

CALCULATED

Example Problem 6 (p. 310):

Tin(II) fluoride, SnF2, is used in some toothpastes. It is made by the reaction of tin with hydrogen fluoride according to the following equation.

Sn(s) + 2 HF(g) SnF2(s) + H2(g) How many grams of SnF2 are produced from the reaction of 30.00 g HF with Sn?

30.00 g HF x

1 mol HF

x 1 mol SnF2 x 156.71 g SnF2 =

20.00 g HF

2 mol HF

1 mol SnF2

117.5 g SnF2

Given

Molar mass from the periodic table

Conversion factor from balanced equation

Calculated answer

Molar mass from the periodic table

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Task 9d

1. When copper metal is added to silver nitrate in solution, silver metal and copper(II) nitrate are produced. What mass of silver is produced from 100.0 g Cu?

2. What mass of aluminum is produced by the decomposition of 5.0 kg Al2O3?

Other conversion factors

Since all of these stoichiometric calculations must go through a mole ratio step, any conversion factor that can be used to change one to another can be use in a problem.

For example: molar volume of a gas

22.4 L of gas at STP mole of gas

Example Problem 7: 32. 7 g of potassium chlorate decomposes to produce potassium chloride and oxygen. What volume of oxygen gas will be produced?

2 KClO3(s) 2 KCl(s) + 3 O2(g)

32.7 g KClO3 x 1 mol KClO3 x

1 mol O2

x

122.6 g KClO3

1 mol KClO3

22.4 L O2 1 mol O2

= 5.97 L O2

Given

Molar mass from the periodic table

Conversion factor from balanced equation

Molar volume of a gas at STP

Calculated answer

Limiting Reactants and Percentage Yield

Limiting Reactants

In the laboratory, a reaction is rarely carried out with exactly the required amount of each of the reactants. In many cases, one or more reactants is present in excess; that is, there is more than the exact amount required to react.

Once one of the reactants is used up, no more products can be formed. The substance that is completely used up first in a reaction is called the limiting reactant. The limiting reactant is the reactant that limits the amount of the other reactant that can combine and the amount of product that can form in a chemical reaction. The substance that is not used up completely in a reaction is called the excess reactant. A limiting reactant may also be referred to as a limiting reagent.

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