Introduction to Stoichiometry

• Stoichiometry: The calculation of reactants and products in chemical reactions.
• Three types of conversions:
  1. Moles of substance A to moles of substance B (single step)
  2. Moles of substance A to grams of substance B or grams of substance A to moles of substance B (two steps)
  3. Grams of substance A to grams of substance B (three steps)

Example Problem 1: Conversion of Moles (SO2 to SO3)

• Chemical Reaction: Sulfur dioxide (SO2) reacts with oxygen (O2) to form sulfur trioxide (SO3).
• Balanced Chemical Equation:
  • SO2 + O2 → SO3
  • Final balanced equation: 2 SO2 + O2 → 2 SO3

Part A

• Given: 3.4 moles of SO2
• Find: Moles of SO3 produced
• Molar Ratio:
  • SO2 : SO3 = 2 : 2 (1:1)
• Calculation:
  • 3.4 moles SO2 × (2 moles SO3 / 2 moles SO2) = 3.4 moles of SO3

Part B

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

Example Problem 2: Conversion of Moles to Grams (Propane to CO2)

• Chemical Reaction: Propane (C3H8) reacts with oxygen to form carbon dioxide (CO2) and water (H2O).
• Balanced Chemical Equation:
  • C3H8 + O2 → CO2 + H2O

Part A

• Given: 2.8 moles of propane
• Find: Grams of CO2 produced
• Steps:
  1. Convert moles of propane to moles of CO2.
     • Molar Ratio: C3H8 : CO2 = 1 : 3
  2. Convert moles of CO2 to grams.
     • Molar Mass of CO2: 44.01 g/mol
• Calculation:
  • Moles of CO2 = 2.8 moles × (3 moles CO2 / 1 mole C3H8) = 8.4 moles of CO2
  • Grams of CO2 = 8.4 moles × 44.01 g/mol = 369.7 g of CO2

Part B

• Given: 3.8 moles of propane
• Find: Grams of O2 required
• Molar Ratio: C3H8 : O2 = 1 : 5
• Molar Mass of O2: 32 g/mol
• Calculation:
  • Moles of O2 = 3.8 moles × (5 moles O2 / 1 mole C3H8) = 19 moles O2
  • Grams of O2 = 19 moles × 32 g/mol = 608 g of O2

Part C

• Given: 25 grams of propane
• Find: Moles of water produced.
• Steps:
  1. Convert grams of propane to moles.
  2. Convert moles of propane to moles of water using the ratio 1:4.
• Molar Mass of Propane (C3H8): 44.094 g/mol
  • Moles of C3H8 = 25 g / 44.094 g/mol = 0.567 moles
  • Moles of H2O = 0.567 moles × 4 = 2.27 moles of H2O

Part D

• Given: 38 grams of water
• Find: Moles of CO2 produced.
• Molar Mass of Water (H2O): 18.016 g/mol
  • Moles of H2O = 38 g / 18.016 g/mol = 2.11 moles
• Molar Ratio: H2O : CO2 = 4 : 3
• Calculation:
  • Moles of CO2 = 2.11 moles × (3 moles CO2 / 4 moles H2O) = 1.58 moles CO2

Example Problem 3: Aluminum and Chlorine Reaction

• Chemical Reaction: Aluminum reacts with chlorine gas to form aluminum chloride.
• Balanced Chemical Equation:
  • 2 Al + 3 Cl2 → 2 AlCl3

Part A

• Given: 35 grams of aluminum
• Find: Grams of aluminum chloride produced.
• Steps:
  1. Convert grams of Al to moles.
  2. Convert moles of Al to moles of AlCl3 using the ratio 2:2.
  3. Convert moles of AlCl3 to grams.
• Molar Mass of Al: 26.98 g/mol
  • Moles of Al = 35 g / 26.98 g/mol = 1.30 moles
  • Moles of AlCl3 = 1.30 moles × (2 moles AlCl3 / 2 moles Al) = 1.30 moles
• Molar Mass of AlCl3: 133.33 g/mol
  • Grams of AlCl3 = 1.30 moles × 133.33 g/mol = 173.00 g of AlCl3

Part B

• Given: 42.8 grams of aluminum
• Find: Grams of chlorine required.
• Steps:
  1. Convert grams of Al to moles.
  2. Convert moles of Al to moles of Cl2 using 2:3 ratio.
  3. Convert moles of Cl2 to grams.
• Calculation:
  • Moles of Cl2 = (42.8 g / 26.98 g/mol) × (3 moles Cl2 / 2 moles Al) × 70.9 g/mol = 168.71 g of Cl2

Conclusion

• Stoichiometry involves various conversions between moles and grams of reactants and products.
• Practice with different types of stoichiometric problems is crucial for mastering the concepts.