Sample Problems For Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry, ensuring that the law of conservation of mass is upheld. This means that the number of atoms of each element must be the same on both the reactant and product sides of the equation. Mastering this skill allows us to accurately predict the quantities of reactants and products involved in a chemical reaction. This article will guide you through various sample problems, providing step-by-step solutions and explanations to help you become proficient in balancing chemical equations.

Introduction to Balancing Chemical Equations

Chemical equations represent chemical reactions using chemical formulas and symbols. A balanced chemical equation shows the exact number of molecules or atoms involved in a reaction, ensuring that matter is neither created nor destroyed.

Why is balancing chemical equations important?

• Conservation of Mass: It adheres to the law of conservation of mass.
• Stoichiometry: It provides the necessary coefficients for stoichiometric calculations, allowing us to determine the amount of reactants needed or products formed in a reaction.
• Accurate Representation: It accurately represents the chemical reaction, showing the correct proportions of reactants and products.

Basic Steps for Balancing Chemical Equations:

1. Write the Unbalanced Equation: Start by writing the chemical formulas for all reactants and products.
2. Count Atoms: Count the number of atoms of each element on both sides of the equation.
3. Balance Elements: Balance the elements one at a time by adjusting coefficients (the numbers in front of the chemical formulas). Start with elements that appear in only one reactant and one product.
4. Check Your Work: After balancing all elements, double-check that the number of atoms of each element is the same on both sides of the equation.
5. Simplify (If Necessary): If all coefficients are divisible by a common number, divide them to obtain the simplest whole-number coefficients.

Sample Problems and Solutions

Let's dive into several examples with detailed solutions to illustrate the process of balancing chemical equations.

Problem 1: Balancing the Combustion of Methane

Unbalanced Equation:

CH₄ + O₂ → CO₂ + H₂O

Step 1: Count Atoms

• Reactant Side:
  • C: 1
  • H: 4
  • O: 2
• Product Side:
  • C: 1
  • H: 2
  • O: 3

Step 2: Balance Hydrogen

To balance hydrogen, we need to multiply H₂O by 2:

CH₄ + O₂ → CO₂ + 2H₂O

Step 3: Count Atoms (Updated)

• Reactant Side:
  • C: 1
  • H: 4
  • O: 2
• Product Side:
  • C: 1
  • H: 4
  • O: 4

Step 4: Balance Oxygen

To balance oxygen, we need to multiply O₂ by 2:

CH₄ + 2O₂ → CO₂ + 2H₂O

Step 5: Count Atoms (Final)

• Reactant Side:
  • C: 1
  • H: 4
  • O: 4
• Product Side:
  • C: 1
  • H: 4
  • O: 4

Balanced Equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

Problem 2: Balancing the Formation of Ammonia

Unbalanced Equation:

N₂ + H₂ → NH₃

Step 1: Count Atoms

• Reactant Side:
  • N: 2
  • H: 2
• Product Side:
  • N: 1
  • H: 3

Step 2: Balance Nitrogen

To balance nitrogen, we need to multiply NH₃ by 2:

N₂ + H₂ → 2NH₃

Step 3: Count Atoms (Updated)

• Reactant Side:
  • N: 2
  • H: 2
• Product Side:
  • N: 2
  • H: 6

Step 4: Balance Hydrogen

To balance hydrogen, we need to multiply H₂ by 3:

N₂ + 3H₂ → 2NH₃

Step 5: Count Atoms (Final)

• Reactant Side:
  • N: 2
  • H: 6
• Product Side:
  • N: 2
  • H: 6

Balanced Equation:

N₂ + 3H₂ → 2NH₃

Problem 3: Balancing the Reaction of Iron with Oxygen

Unbalanced Equation:

Fe + O₂ → Fe₂O₃

Step 1: Count Atoms

• Reactant Side:
  • Fe: 1
  • O: 2
• Product Side:
  • Fe: 2
  • O: 3

Step 2: Balance Iron

To balance iron, we need to multiply Fe by 2:

2Fe + O₂ → Fe₂O₃

Step 3: Count Atoms (Updated)

• Reactant Side:
  • Fe: 2
  • O: 2
• Product Side:
  • Fe: 2
  • O: 3

Step 4: Balance Oxygen

To balance oxygen, we need to find a common multiple for 2 and 3, which is 6. Multiply O₂ by 3 and Fe₂O₃ by 2:

2Fe + 3O₂ → 2Fe₂O₃

Step 5: Count Atoms (Updated)

• Reactant Side:
  • Fe: 2
  • O: 6
• Product Side:
  • Fe: 4
  • O: 6

Step 6: Balance Iron (Again)

Now we need to balance iron again. Multiply Fe by 4:

4Fe + 3O₂ → 2Fe₂O₃

Step 7: Count Atoms (Final)

• Reactant Side:
  • Fe: 4
  • O: 6
• Product Side:
  • Fe: 4
  • O: 6

Balanced Equation:

4Fe + 3O₂ → 2Fe₂O₃

Problem 4: Balancing the Reaction of Potassium with Water

Unbalanced Equation:

K + H₂O → KOH + H₂

Step 1: Count Atoms

• Reactant Side:
  • K: 1
  • H: 2
  • O: 1
• Product Side:
  • K: 1
  • H: 3
  • O: 1

Step 2: Balance Hydrogen

To balance hydrogen, we need to multiply H₂O by 2 and KOH by 2:

K + 2H₂O → 2KOH + H₂

Step 3: Count Atoms (Updated)

• Reactant Side:
  • K: 1
  • H: 4
  • O: 2
• Product Side:
  • K: 2
  • H: 3
  • O: 2

Step 4: Balance Potassium

To balance potassium, multiply K by 2:

2K + 2H₂O → 2KOH + H₂

Step 5: Count Atoms (Final)

• Reactant Side:
  • K: 2
  • H: 4
  • O: 2
• Product Side:
  • K: 2
  • H: 4
  • O: 2

Balanced Equation:

2K + 2H₂O → 2KOH + H₂

Problem 5: Balancing the Reaction of Aluminum with Hydrochloric Acid

Unbalanced Equation:

Al + HCl → AlCl₃ + H₂

Step 1: Count Atoms

• Reactant Side:
  • Al: 1
  • H: 1
  • Cl: 1
• Product Side:
  • Al: 1
  • H: 2
  • Cl: 3

Step 2: Balance Chlorine

To balance chlorine, multiply HCl by 3:

Al + 3HCl → AlCl₃ + H₂

Step 3: Count Atoms (Updated)

• Reactant Side:
  • Al: 1
  • H: 3
  • Cl: 3
• Product Side:
  • Al: 1
  • H: 2
  • Cl: 3

Step 4: Balance Hydrogen

To balance hydrogen, find a common multiple for 3 and 2, which is 6. Multiply HCl by 6 and H₂ by 3:

Al + 6HCl → AlCl₃ + 3H₂

Step 5: Count Atoms (Updated)

• Reactant Side:
  • Al: 1
  • H: 6
  • Cl: 6
• Product Side:
  • Al: 1
  • H: 6
  • Cl: 3

Step 6: Balance Chlorine (Again)

To balance chlorine, multiply AlCl₃ by 2:

Al + 6HCl → 2AlCl₃ + 3H₂

Step 7: Count Atoms (Updated)

• Reactant Side:
  • Al: 1
  • H: 6
  • Cl: 6
• Product Side:
  • Al: 2
  • H: 6
  • Cl: 6

Step 8: Balance Aluminum

To balance aluminum, multiply Al by 2:

2Al + 6HCl → 2AlCl₃ + 3H₂

Step 9: Count Atoms (Final)

• Reactant Side:
  • Al: 2
  • H: 6
  • Cl: 6
• Product Side:
  • Al: 2
  • H: 6
  • Cl: 6

Balanced Equation:

2Al + 6HCl → 2AlCl₃ + 3H₂

Problem 6: Balancing the Reaction of Copper with Nitric Acid

This is a more complex reaction involving polyatomic ions and redox processes.

Unbalanced Equation:

Cu + HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O

Step 1: Count Atoms

• Reactant Side:
  • Cu: 1
  • H: 1
  • N: 1
  • O: 3
• Product Side:
  • Cu: 1
  • H: 2
  • N: 3 (2 from Cu(NO₃)₂, 1 from NO₂)
  • O: 9 (6 from Cu(NO₃)₂, 2 from NO₂, 1 from H₂O)

Step 2: Balance Copper

Copper is already balanced, so no change is needed.

Step 3: Balance Nitrogen

The nitrogen is a bit tricky because it appears in two different products. Let's start by balancing the nitrogen in Cu(NO₃)₂. Multiply HNO₃ by 2 to account for the two nitrate ions:

Cu + 4HNO₃ → Cu(NO₃)₂ + NO₂ + H₂O

Step 4: Count Atoms (Updated)

• Reactant Side:
  • Cu: 1
  • H: 4
  • N: 4
  • O: 12
• Product Side:
  • Cu: 1
  • H: 2
  • N: 3
  • O: 9

Step 5: Balance Nitrogen (Again)

Now, let's balance the remaining nitrogen by multiplying NO₂ by 2:

Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + H₂O

Step 6: Count Atoms (Updated)

• Reactant Side:
  • Cu: 1
  • H: 4
  • N: 4
  • O: 12
• Product Side:
  • Cu: 1
  • H: 2
  • N: 4
  • O: 11

Step 7: Balance Hydrogen

Balance hydrogen by multiplying H₂O by 2:

Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

Step 8: Count Atoms (Final)

• Reactant Side:
  • Cu: 1
  • H: 4
  • N: 4
  • O: 12
• Product Side:
  • Cu: 1
  • H: 4
  • N: 4
  • O: 12

Balanced Equation:

Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

Problem 7: Balancing a Redox Reaction in Acidic Solution (Ion-Electron Method)

This method is used for balancing redox reactions, especially those in acidic or basic solutions.

Unbalanced Equation (Skeleton Equation):

MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺  (in acidic solution)

Step 1: Write Half-Reactions

• Oxidation Half-Reaction: Fe²⁺ → Fe³⁺
• Reduction Half-Reaction: MnO₄⁻ → Mn²⁺

Step 2: Balance Atoms (Except O and H)

• Oxidation Half-Reaction: Fe²⁺ → Fe³⁺ (Already balanced)
• Reduction Half-Reaction: MnO₄⁻ → Mn²⁺ (Manganese is balanced)

Step 3: Balance Oxygen by Adding H₂O

• Oxidation Half-Reaction: Fe²⁺ → Fe³⁺ (No oxygen to balance)
• Reduction Half-Reaction: MnO₄⁻ → Mn²⁺ + 4H₂O

Step 4: Balance Hydrogen by Adding H⁺

• Oxidation Half-Reaction: Fe²⁺ → Fe³⁺ (No hydrogen to balance)
• Reduction Half-Reaction: 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O

Step 5: Balance Charge by Adding Electrons (e⁻)

• Oxidation Half-Reaction: Fe²⁺ → Fe³⁺ + e⁻ (Add 1 electron to the right)
• Reduction Half-Reaction: 5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O (Add 5 electrons to the left)

Step 6: Make the Number of Electrons Equal in Both Half-Reactions

• Multiply the oxidation half-reaction by 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻
• Reduction Half-Reaction: 5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O

Step 7: Add the Half-Reactions and Cancel Out Electrons

5Fe²⁺ → 5Fe³⁺ + 5e⁻
5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O
-----------------------------------------
5Fe²⁺ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Step 8: Check that Atoms and Charges are Balanced

• Atoms:
  • Fe: 5 on both sides
  • Mn: 1 on both sides
  • O: 4 on both sides
  • H: 8 on both sides
• Charge:
  • Reactant Side: (5 x +2) + (8 x +1) + (-1) = +17
  • Product Side: (5 x +3) + (+2) = +17

Balanced Equation:

5Fe²⁺ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Problem 8: Balancing a Redox Reaction in Basic Solution (Ion-Electron Method)

Unbalanced Equation (Skeleton Equation):

Cl₂ + OH⁻ → Cl⁻ + ClO₃⁻  (in basic solution)

Step 1: Write Half-Reactions

• Reduction Half-Reaction: Cl₂ → Cl⁻
• Oxidation Half-Reaction: Cl₂ → ClO₃⁻

Step 2: Balance Atoms (Except O and H)

• Reduction Half-Reaction: Cl₂ → 2Cl⁻
• Oxidation Half-Reaction: Cl₂ → 2ClO₃⁻

Step 3: Balance Oxygen by Adding H₂O

• Reduction Half-Reaction: Cl₂ → 2Cl⁻ (No oxygen to balance)
• Oxidation Half-Reaction: Cl₂ + 6H₂O → 2ClO₃⁻

Step 4: Balance Hydrogen by Adding H⁺

• Reduction Half-Reaction: Cl₂ → 2Cl⁻ (No hydrogen to balance)
• Oxidation Half-Reaction: Cl₂ + 6H₂O → 2ClO₃⁻ + 12H⁺

Step 5: Balance Charge by Adding Electrons (e⁻)

• Reduction Half-Reaction: Cl₂ + 2e⁻ → 2Cl⁻
• Oxidation Half-Reaction: Cl₂ + 6H₂O → 2ClO₃⁻ + 12H⁺ + 10e⁻

Step 6: Make the Number of Electrons Equal in Both Half-Reactions

• Multiply the reduction half-reaction by 5: 5Cl₂ + 10e⁻ → 10Cl⁻
• Oxidation Half-Reaction: Cl₂ + 6H₂O → 2ClO₃⁻ + 12H⁺ + 10e⁻

Step 7: Add the Half-Reactions and Cancel Out Electrons

5Cl₂ + 10e⁻ → 10Cl⁻
Cl₂ + 6H₂O → 2ClO₃⁻ + 12H⁺ + 10e⁻
-----------------------------------------
6Cl₂ + 6H₂O → 10Cl⁻ + 2ClO₃⁻ + 12H⁺

Step 8: Convert to Basic Solution by Adding OH⁻ to Neutralize H⁺

Add 12OH⁻ to both sides to neutralize the 12H⁺:

6Cl₂ + 6H₂O + 12OH⁻ → 10Cl⁻ + 2ClO₃⁻ + 12H⁺ + 12OH⁻

The 12H⁺ and 12OH⁻ combine to form 12H₂O:

6Cl₂ + 6H₂O + 12OH⁻ → 10Cl⁻ + 2ClO₃⁻ + 12H₂O

Step 9: Simplify by Canceling Out Water Molecules

6Cl₂ + 12OH⁻ → 10Cl⁻ + 2ClO₃⁻ + 6H₂O

Step 10: Simplify Coefficients (If Possible)

Divide all coefficients by 2:

3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O

Step 11: Check that Atoms and Charges are Balanced

• Atoms:
  • Cl: 6 on both sides
  • O: 6 on both sides
  • H: 6 on both sides
• Charge:
  • Reactant Side: (6 x -1) = -6
  • Product Side: (5 x -1) + (-1) = -6

Balanced Equation:

3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O

Tips and Tricks for Balancing Chemical Equations

• Start with the Most Complex Molecule: Begin balancing with the molecule that has the most atoms or the most different elements.
• Balance Polyatomic Ions as a Unit: If a polyatomic ion appears on both sides of the equation unchanged, treat it as a single unit.
• Leave Hydrogen and Oxygen for Last: Balance hydrogen and oxygen after all other elements.
• Use Fractions: If necessary, use fractions to balance elements, but remember to multiply the entire equation by the denominator to get whole-number coefficients.
• Practice Regularly: The more you practice, the more comfortable and efficient you will become at balancing chemical equations.

Common Mistakes to Avoid

• Changing Subscripts: Only change coefficients, never subscripts. Changing subscripts changes the chemical formula of the substance.
• Forgetting to Distribute Coefficients: Ensure that the coefficient applies to all atoms within the molecule or ion.
• Not Checking Your Work: Always double-check that the number of atoms of each element is the same on both sides of the equation.
• Stopping Too Early: Make sure the coefficients are in the simplest whole-number ratio.

Conclusion

Balancing chemical equations is an essential skill in chemistry that ensures the conservation of mass and provides accurate stoichiometric relationships. By following a systematic approach and practicing with various examples, you can master this skill. Remember to count atoms, balance elements one at a time, check your work, and simplify coefficients when necessary. With consistent effort and attention to detail, you'll be able to confidently balance even the most complex chemical equations. The sample problems and solutions provided in this article should serve as a valuable resource for your learning journey. Keep practicing, and you'll find that balancing chemical equations becomes second nature.