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Introductory Chemistry - 1st Canadian Edition: The Mole in Chemical Reactions

Introductory Chemistry - 1st Canadian Edition
The Mole in Chemical Reactions
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table of contents
  1. Cover
  2. Title Page
  3. Copyright
  4. Table Of Contents
  5. Acknowledgments
  6. Dedication
  7. About BCcampus Open Education
  8. Chapter 1. What is Chemistry
    1. Some Basic Definitions
    2. Chemistry as a Science
  9. Chapter 2. Measurements
    1. Expressing Numbers
    2. Significant Figures
    3. Converting Units
    4. Other Units: Temperature and Density
    5. Expressing Units
    6. End-of-Chapter Material
  10. Chapter 3. Atoms, Molecules, and Ions
    1. Acids
    2. Ions and Ionic Compounds
    3. Masses of Atoms and Molecules
    4. Molecules and Chemical Nomenclature
    5. Atomic Theory
    6. End-of-Chapter Material
  11. Chapter 4. Chemical Reactions and Equations
    1. The Chemical Equation
    2. Types of Chemical Reactions: Single- and Double-Displacement Reactions
    3. Ionic Equations: A Closer Look
    4. Composition, Decomposition, and Combustion Reactions
    5. Oxidation-Reduction Reactions
    6. Neutralization Reactions
    7. End-of-Chapter Material
  12. Chapter 5. Stoichiometry and the Mole
    1. Stoichiometry
    2. The Mole
    3. Mole-Mass and Mass-Mass Calculations
    4. Limiting Reagents
    5. The Mole in Chemical Reactions
    6. Yields
    7. End-of-Chapter Material
  13. Chapter 6. Gases
    1. Pressure
    2. Gas Laws
    3. Other Gas Laws
    4. The Ideal Gas Law and Some Applications
    5. Gas Mixtures
    6. Kinetic Molecular Theory of Gases
    7. Molecular Effusion and Diffusion
    8. Real Gases
    9. End-of-Chapter Material
  14. Chapter 7. Energy and Chemistry
    1. Formation Reactions
    2. Energy
    3. Stoichiometry Calculations Using Enthalpy
    4. Enthalpy and Chemical Reactions
    5. Work and Heat
    6. Hess’s Law
    7. End-of-Chapter Material
  15. Chapter 8. Electronic Structure
    1. Light
    2. Quantum Numbers for Electrons
    3. Organization of Electrons in Atoms
    4. Electronic Structure and the Periodic Table
    5. Periodic Trends
    6. End-of-Chapter Material
  16. Chapter 9. Chemical Bonds
    1. Lewis Electron Dot Diagrams
    2. Electron Transfer: Ionic Bonds
    3. Covalent Bonds
    4. Other Aspects of Covalent Bonds
    5. Violations of the Octet Rule
    6. Molecular Shapes and Polarity
    7. Valence Bond Theory and Hybrid Orbitals
    8. Molecular Orbitals
    9. End-of-Chapter Material
  17. Chapter 10. Solids and Liquids
    1. Properties of Liquids
    2. Solids
    3. Phase Transitions: Melting, Boiling, and Subliming
    4. Intermolecular Forces
    5. End-of-Chapter Material
  18. Chapter 11. Solutions
    1. Colligative Properties of Solutions
    2. Concentrations as Conversion Factors
    3. Quantitative Units of Concentration
    4. Colligative Properties of Ionic Solutes
    5. Some Definitions
    6. Dilutions and Concentrations
    7. End-of-Chapter Material
  19. Chapter 12. Acids and Bases
    1. Acid-Base Titrations
    2. Strong and Weak Acids and Bases and Their Salts
    3. Brønsted-Lowry Acids and Bases
    4. Arrhenius Acids and Bases
    5. Autoionization of Water
    6. Buffers
    7. The pH Scale
    8. End-of-Chapter Material
  20. Chapter 13. Chemical Equilibrium
    1. Chemical Equilibrium
    2. The Equilibrium Constant
    3. Shifting Equilibria: Le Chatelier’s Principle
    4. Calculating Equilibrium Constant Values
    5. Some Special Types of Equilibria
    6. End-of-Chapter Material
  21. Chapter 14. Oxidation and Reduction
    1. Oxidation-Reduction Reactions
    2. Balancing Redox Reactions
    3. Applications of Redox Reactions: Voltaic Cells
    4. Electrolysis
    5. End-of-Chapter Material
  22. Chapter 15. Nuclear Chemistry
    1. Units of Radioactivity
    2. Uses of Radioactive Isotopes
    3. Half-Life
    4. Radioactivity
    5. Nuclear Energy
    6. End-of-Chapter Material
  23. Chapter 16. Organic Chemistry
    1. Hydrocarbons
    2. Branched Hydrocarbons
    3. Alkyl Halides and Alcohols
    4. Other Oxygen-Containing Functional Groups
    5. Other Functional Groups
    6. Polymers
    7. End-of-Chapter Material
  24. Chapter 17. Kinetics
    1. Factors that Affect the Rate of Reactions
    2. Reaction Rates
    3. Rate Laws
    4. Concentration–Time Relationships: Integrated Rate Laws
    5. Activation Energy and the Arrhenius Equation
    6. Reaction Mechanisms
    7. Catalysis
    8. End-of-Chapter Material
  25. Chapter 18. Chemical Thermodynamics
    1. Spontaneous Change
    2. Entropy and the Second Law of Thermodynamics
    3. Measuring Entropy and Entropy Changes
    4. Gibbs Free Energy
    5. Spontaneity: Free Energy and Temperature
    6. Free Energy under Nonstandard Conditions
    7. End-of-Chapter Material
  26. Appendix A: Periodic Table of the Elements
  27. Appendix B: Selected Acid Dissociation Constants at 25°C
  28. Appendix C: Solubility Constants for Compounds at 25°C
  29. Appendix D: Standard Thermodynamic Quantities for Chemical Substances at 25°C
  30. Appendix E: Standard Reduction Potentials by Value
  31. Glossary
  32. About the Authors
  33. Versioning History

The Mole in Chemical Reactions

Learning Objectives

  1. Balance a chemical equation in terms of moles.
  2. Use the balanced equation to construct conversion factors in terms of moles.
  3. Calculate moles of one substance from moles of another substance using a balanced chemical equation.

Consider this balanced chemical equation:

2H2 + O2 → 2H2O

We interpret this as “two molecules of hydrogen react with one molecule of oxygen to make two molecules of water.” The chemical equation is balanced as long as the coefficients are in the ratio 2:1:2. For instance, this chemical equation is also balanced:

100H2 + 50O2 → 100H2O

This equation is not conventional—because convention says that we use the lowest ratio of coefficients — but it is balanced. So is this chemical equation:

5,000H2 + 2,500O2 → 5,000H2O

Again, this is not conventional, but it is still balanced. Suppose we use a much larger number:

12.044 × 1023 H2 + 6.022 × 1023 O2 → 12.044 × 1023 H2O

These coefficients are also in the ratio of 2:1:2. But these numbers are related to the number of things in a mole: the first and last numbers are two times Avogadro’s number, while the second number is Avogadro’s number. That means that the first and last numbers represent 2 mol, while the middle number is just 1 mol. Well, why not just use the number of moles in balancing the chemical equation?

2H2 + O2 → 2H2O

This is the same balanced chemical equation we started with! What this means is that chemical equations are not just balanced in terms of molecules; they are also balanced in terms of moles. We can just as easily read this chemical equation as “two moles of hydrogen react with one mole of oxygen to make two moles of water.” All balanced chemical reactions are balanced in terms of moles.

Example 5.12

Problem

Interpret this balanced chemical equation in terms of moles.

P4 + 5O2 → P4O10

Solution

The coefficients represent the number of moles that react, not just molecules. We would speak of this equation as “one mole of molecular phosphorus reacts with five moles of elemental oxygen to make one mole of tetraphosphorus decoxide.”

Test Yourself

Interpret this balanced chemical equation in terms of moles.

N2 + 3H2 → 2NH3

Answer

One mole of elemental nitrogen reacts with three moles of elemental hydrogen to produce two moles of ammonia.

In Chapter 4 “Chemical Reactions and Equations”, in the section called “The Chemical Equation”, we stated that a chemical equation is simply a recipe for a chemical reaction. As such, chemical equations also give us equivalences—equivalences between the reactants and the products. However, now we understand that these equivalences are expressed in terms of moles. Consider the following chemical equation:

2H2 + O2 → 2H2O

This chemical reaction gives us the following equivalences:

2 mol H2 ⇔ 1 mol O2 ⇔ 2 mol H2O

Any two of these quantities can be used to construct a conversion factor that lets us relate the number of moles of one substance to an equivalent number of moles of another substance. If, for example, we want to know how many moles of oxygen will react with 17.6 mol of hydrogen, we construct a conversion factor between 2 mol of H2 and 1 mol of O2 and use it to convert from moles of one substance to moles of another:

17.6\cancel{\text{ mol }\ce{H2}}\times \dfrac{1\text{ mol }\ce{O2}}{2\cancel{\text{ mol }\ce{H2}}}=8.80\text{ mol }\ce{O2}

Note how the mol H2 unit cancels, and mol O2 is the new unit introduced. This is an example of a mole-mole calculation, when you start with moles of one substance and convert to moles of another substance by using the balanced chemical equation. The example may seem simple because the numbers are small, but numbers won’t always be so simple!

Example 5.13

Problem

Consider the following balanced chemical equation:

2C4H10(g) + 13O2 → 8CO2(g) + 10H2O(ℓ)

If 154 mol of O2 are reacted, how many moles of CO2 are produced?

Solution

We are relating an amount of oxygen to an amount of carbon dioxide, so we need the equivalence between these two substances. According to the balanced chemical equation, the equivalence is:

13 mol O2 ⇔ 8 mol CO2

We can use this equivalence to construct the proper conversion factor. We start with what we are given and apply the conversion factor:

154\cancel{\text{ mol }\ce{O2}}\times \dfrac{8\text{ mol }\ce{CO2}}{13\cancel{\text{ mol }\ce{O2}}}=94.8\text{ mol }\ce{CO2}

The mol O2 unit is in the denominator of the conversion factor so it cancels. Both the 8 and the 13 are exact numbers, so they don’t contribute to the number of significant figures in the final answer.

Test Yourself

Using the above equation, how many moles of H2O are produced when 154 mol of O2 react?

Answer

118 mol

It is important to reiterate that balanced chemical equations are balanced in terms of moles. Not grams, kilograms, or litres — but moles. Any stoichiometry problem will likely need to work through the mole unit at some point, especially if you are working with a balanced chemical reaction.

Key Takeaways

  • Balanced chemical reactions are balanced in terms of moles.
  • A balanced chemical reaction gives equivalences in moles that allow stoichiometry calculations to be performed.

Exercises

Questions

  1. Express in mole terms what this chemical equation means.

    CH4 + 2O2 → CO2 + 2H2O

  2. Express in mole terms what this chemical equation means.

    Na2CO3 + 2HCl → 2NaCl + H2O + CO2

  3. How many molecules of each substance are involved in the equation in Exercise 1 if it is interpreted in terms of moles?
  4. How many molecules of each substance are involved in the equation in Exercise 2 if it is interpreted in terms of moles?
  5. For the chemical equation 2C2H6 + 7O2 → 4CO2 + 6H2O, what equivalences can you write in terms of moles? Use the ⇔ sign.
  6. For the chemical equation 2Al + 3Cl2 → 2AlCl3, what equivalences can you write in terms of moles? Use the ⇔ sign.
  7. Write the balanced chemical reaction for the combustion of C5H12 (the products are CO2 and H2O) and determine how many moles of H2O are formed when 5.8 mol of O2 are reacted.
  8. Write the balanced chemical reaction for the formation of Fe2(SO4)3 from Fe2O3 and SO3 and determine how many moles of Fe2(SO4)3 are formed when 12.7 mol of SO3 are reacted.
  9. For the balanced chemical equation 3Cu(s) + 2NO3−(aq) + 8H+(aq) → 3Cu2+(aq) + 4H2O(ℓ) + 2NO(g), how many moles of Cu2+ are formed when 55.7 mol of H+ are reacted?
  10. For the balanced chemical equation Al(s) + 3Ag+(aq) → Al3+(aq) + 3Ag(s), how many moles of Ag are produced when 0.661 mol of Al are reacted?
  11. For the balanced chemical reaction 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(ℓ), how many moles of H2O are produced when 0.669 mol of NH3 react?
  12. For the balanced chemical reaction 4NaOH(aq) + 2S(s) + 3O2(g) → 2Na2SO4(aq) + 2H2O(ℓ), how many moles of Na2SO4 are formed when 1.22 mol of O2 react?
  13. For the balanced chemical reaction 4KO2(s) + 2CO2(g) → 2K2CO3(s) + 3O2(g), determine the number of moles of both products formed when 6.88 mol of KO2 react.
  14. For the balanced chemical reaction 2AlCl3 + 3H2O(ℓ) → Al2O3 + 6HCl(g), determine the number of moles of both products formed when 0.0552 mol of AlCl3 react.

Answers

  1. One mole of CH4 reacts with 2 mol of O2 to make 1 mol of CO2 and 2 mol of H2O.
  1. 6.022 × 1023 molecules of CH4, 1.2044 × 1024 molecules of O2, 6.022 × 1023 molecules of CO2, and 1.2044 × 1024 molecules of H2O
  1. 2 mol of C2H6 ⇔ 7 mol of O2 ⇔ 4 mol of CO2 ⇔ 6 mol of H2O
  1. C5H12 + 8 O2 → 5CO2 + 6H2O; 4.4 mol
  1. 20.9 mol
  1. 1.00 mol
  1. 3.44 mol of K2CO3; 5.16 mol of O2

Annotate

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Copyright © 2014

                                by Jessie A. Key

            Introductory Chemistry - 1st Canadian Edition by Jessie A. Key is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.
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