Stoichiometry in Combustion Reactions

Introduction to Combustion

• Combustion is a chemical process involving rapid reaction between a substance with an oxidant, usually oxygen, to produce heat and light.
• Fuels react with oxygen to produce oxides (e.g., \(CO_2\), \(H_2O\)).
• Complete combustion requires sufficient oxygen; incomplete combustion occurs when oxygen is limited, producing carbon monoxide (CO) and/or carbon (C) as byproducts.
• Thermochemical equations include the enthalpy change (\(\Delta H\)) of the reaction. For combustion, \(\Delta H\) is negative (exothermic).

KEY TAKEAWAY: Combustion is an exothermic reaction that releases heat. Complete combustion produces \(CO_2\) and \(H_2O\), while incomplete combustion produces \(CO\), \(C\), and \(H_2O\).

Stoichiometry Principles

• Stoichiometry is the study of the quantitative relationships or ratios between two or more substances undergoing a physical change or chemical reaction.
• It is based on the law of conservation of mass.
• Mole ratio: The ratio of the amounts in moles of any two substances involved in a chemical reaction.

Steps for Stoichiometric Calculations:

1. Write a balanced chemical equation.
2. Convert the given mass or volume of the known substance to moles using the molar mass or molar volume (at SLC).
3. Use the mole ratio from the balanced equation to determine the moles of the unknown substance.
4. Convert the moles of the unknown substance to the desired units (mass, volume, etc.).

REMEMBER: Moles are central to stoichiometric calculations. Use the formula: \(n = \frac{m}{M}\) (where \(n\) = moles, \(m\) = mass, \(M\) = molar mass). At SLC, \(n = \frac{V}{V_m}\) (where \(V\) = volume and \(V_m\)= 24.8 L/mol).

Combustion Stoichiometry Calculations

1. Calculating Mass of Reactants/Products

• Example: Calculate the mass of \(CO_2\) produced when 10.0 g of methane (\(CH_4\)) undergoes complete combustion.

  1. Balanced equation: \(CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)\)
  2. Moles of \(CH_4\): \(n(CH_4) = \frac{10.0 \ g}{16.04 \ g/mol} = 0.623 \ mol\)
  3. Mole ratio \(CH_4 : CO_2\) is 1:1. Therefore, \(n(CO_2) = 0.623 \ mol\)
  4. Mass of \(CO_2\): \(m(CO_2) = 0.623 \ mol \times 44.01 \ g/mol = 27.4 \ g\)

2. Calculating Volume of Gaseous Reactants/Products (at SLC)

• Standard laboratory conditions (SLC): 25°C and 100 kPa. Molar volume (\(V_m\)) at SLC is 24.8 L/mol.

• Example: What volume of oxygen (at SLC) is required for the complete combustion of 5.0 g of ethane (\(C_2H_6\))?

  1. Balanced equation: \(2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)\)
  2. Moles of \(C_2H_6\): \(n(C_2H_6) = \frac{5.0 \ g}{30.07 \ g/mol} = 0.166 \ mol\)
  3. Mole ratio \(C_2H_6 : O_2\) is 2:7. Therefore, \(n(O_2) = 0.166 \ mol \times \frac{7}{2} = 0.581 \ mol\)
  4. Volume of \(O_2\) at SLC: \(V(O_2) = 0.581 \ mol \times 24.8 \ L/mol = 14.4 \ L\)

3. Calculating Energy Released

• Use the thermochemical equation to determine the energy released per mole of fuel.

• Example: Given the thermochemical equation for methane combustion: \(CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \ \ \Delta H = -890 \ kJ/mol\). Calculate the energy released when 3.2 g of methane is burned.

  1. Moles of \(CH_4\): \(n(CH_4) = \frac{3.2 \ g}{16.04 \ g/mol} = 0.20 \ mol\)
  2. Energy released: \(Energy = 0.20 \ mol \times 890 \ kJ/mol = 178 \ kJ\)

4. Limiting Reactant and Excess Reactant

• The limiting reactant is the reactant that is completely consumed in a reaction and determines the amount of product formed.
• The excess reactant is the reactant present in a greater amount than necessary to react completely with the limiting reactant.

• To determine the limiting reactant:

  1. Calculate the moles of each reactant.
  2. Divide the moles of each reactant by its stoichiometric coefficient in the balanced equation.
  3. The reactant with the smallest value is the limiting reactant.

• Example: 5.0 g of methane (\(CH_4\)) is reacted with 10.0 g of oxygen (\(O_2\)). Determine the limiting reactant.

  1. Balanced equation: \(CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)\)
  2. Moles of \(CH_4\): \(n(CH_4) = \frac{5.0 \ g}{16.04 \ g/mol} = 0.312 \ mol\)
  3. Moles of \(O_2\): \(n(O_2) = \frac{10.0 \ g}{32.00 \ g/mol} = 0.313 \ mol\)
  4. \(CH_4\): \(\frac{0.312 \ mol}{1} = 0.312\)
  5. \(O_2\): \(\frac{0.313 \ mol}{2} = 0.157\)
  6. Since 0.157 < 0.312, \(O_2\) is the limiting reactant.

EXAM TIP: Always write the balanced equation first! Pay attention to units and ensure they are consistent (e.g., grams to moles). When asked to calculate the amount of product formed, base your calculation on the limiting reactant.

Energy Transformation Efficiency

• Energy transformation efficiency is the percentage of chemical energy converted to useful energy.
• \[Efficiency = \frac{Useful \ Energy \ Output}{Total \ Energy \ Input} \times 100\%\]

• Example: A combustion reaction releases 500 kJ of energy, but only 350 kJ is used to heat water. Calculate the energy transformation efficiency.

  • \(Efficiency = \frac{350 \ kJ}{500 \ kJ} \times 100\% = 70\%\)

Energy Values of Foods

• Foods contain carbohydrates, proteins, and fats/oils, which provide energy when combusted (metabolized) in the body.
• Energy values are typically expressed in kJ/g.

• Approximate energy values:

  • Carbohydrates: 17 kJ/g
  • Proteins: 17 kJ/g
  • Fats/Oils: 37 kJ/g

• Example: Calculate the total energy content of a food containing 10 g of carbohydrates, 5 g of protein, and 3 g of fat.

  • Energy from carbohydrates: \(10 \ g \times 17 \ kJ/g = 170 \ kJ\)
  • Energy from proteins: \(5 \ g \times 17 \ kJ/g = 85 \ kJ\)
  • Energy from fats: \(3 \ g \times 37 \ kJ/g = 111 \ kJ\)
  • Total energy: \(170 \ kJ + 85 \ kJ + 111 \ kJ = 366 \ kJ\)

COMMON MISTAKE: Forgetting to balance the combustion equation before performing stoichiometric calculations. Double-check your work!

Solution Calorimetry (brief mention, as per Textbook Context)

• Solution calorimetry involves measuring the heat change when a substance dissolves in a solvent.
• A solution calorimeter is used to measure the heat released or absorbed during a chemical reaction in solution.
• Calibration factor is determined by introducing a known amount of heat into the calorimeter and measuring the temperature change.
• Account for heat loss to the surroundings for accurate measurements.

VCAA FOCUS: VCAA often includes questions relating to calculating masses, volumes, and energy released from combustion reactions. Practice a variety of problems.