Introduction to Stoichiometry

• Definition: Stoichiometry involves the calculation of reactants and products in chemical reactions.
• Types of Conversions:
  1. Moles of Substance A to Moles of Substance B (1-step)
  2. Moles of Substance A to Grams of Substance B or Grams of Substance A to Moles of Substance B (2-steps)
  3. Grams of Substance A to Grams of Substance B (3-steps)

Example Problem 1: Sulfur Dioxide Reaction

Reaction Details

• Chemical Reaction: SO2 + O2 → SO3
• Balanced Equation: 2SO2 + O2 → 2SO3

Part A: Moles of SO3 from SO2

• Given: 3.4 moles of SO2
• Molar Ratio: 2 moles SO2 : 2 moles SO3
• Calculation:
  • Moles of SO3 = 3.4 moles SO2 × (2 moles SO3 / 2 moles SO2) = 3.4 moles SO3

Part B: Moles of O2 from SO2

• Given: 4.7 moles of SO2
• Molar Ratio: 2 moles SO2 : 1 mole O2
• Calculation:
  • Moles of O2 = 4.7 moles SO2 × (1 mole O2 / 2 moles SO2) = 2.35 moles O2

Example Problem 2: Propane Reaction

Reaction Details

• Chemical Reaction: C3H8 + O2 → CO2 + H2O
• Balanced Equation: 1C3H8 + 5O2 → 3CO2 + 4H2O

Part A: Grams of CO2 from Propane

• Given: 2.8 moles of C3H8
• Molar Ratio: 1 mole C3H8 : 3 moles CO2
• Molar Mass of CO2: 44.01 g/mol
• Calculation:
  • Grams of CO2 = 2.8 moles C3H8 × (3 moles CO2 / 1 mole C3H8) × 44.01 g/mol = 369.7 grams CO2

Part B: Grams of O2 from Propane

• Given: 3.8 moles of C3H8
• Molar Ratio: 1 mole C3H8 : 5 moles O2
• Molar Mass of O2: 32 g/mol
• Calculation:
  • Grams of O2 = 3.8 moles C3H8 × (5 moles O2 / 1 mole C3H8) × 32 g/mol = 608 grams O2

Part C: Moles of H2O from Grams of C3H8

• Given: 25 grams of C3H8
• Molar Mass of C3H8: 44.094 g/mol
• Calculation Steps:
  1. Convert grams to moles: 25 g / 44.094 g/mol = 0.567 moles C3H8
  2. Use Molar Ratio: 0.567 moles C3H8 × (4 moles H2O / 1 mole C3H8) = 2.27 moles H2O

Part D: Moles of CO2 from Grams of H2O

• Given: 38 grams of H2O
• Molar Mass of H2O: 18.016 g/mol
• Calculation Steps:
  1. Convert grams to moles: 38 g / 18.016 g/mol = 2.11 moles H2O
  2. Use Molar Ratio: 2.11 moles H2O × (3 moles CO2 / 4 moles H2O) = 1.58 moles CO2

Example Problem 3: Aluminum Reaction

Reaction Details

• Chemical Reaction: Al + Cl2 → AlCl3
• Balanced Equation: 2Al + 3Cl2 → 2AlCl3

Part A: Grams of AlCl3 from Grams of Al

• Given: 35 grams of Al
• Molar Mass of Al: 26.98 g/mol
• Calculation Steps:
  1. Convert grams to moles: 35 g / 26.98 g/mol = 1.30 moles Al
  2. Use Molar Ratio: 1.30 moles Al × (2 moles AlCl3 / 2 moles Al) = 1.30 moles AlCl3
  3. Molar Mass of AlCl3: 133.33 g/mol
  4. Convert moles to grams: 1.30 moles AlCl3 × 133.33 g/mol = 172.96 grams AlCl3

Part B: Grams of Cl2 from Grams of Al

• Given: 42.8 grams of Al
• Calculation Steps:
  1. Convert grams to moles: 42.8 g / 26.98 g/mol = 1.59 moles Al
  2. Use Molar Ratio: 1.59 moles Al × (3 moles Cl2 / 2 moles Al) = 2.39 moles Cl2
  3. Molar Mass of Cl2: 70.9 g/mol
  4. Convert moles to grams: 2.39 moles Cl2 × 70.9 g/mol = 168.71 grams Cl2

Conclusion

• Key Points:
  • Understand mole-to-mole conversions, grams to moles conversions, and gram-to-gram conversions.
  • Practice with balanced equations is crucial for stoichiometric calculations.