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Percent Yield: It is the ratio of the actual yield to the theoretical yield expressed as a percentage.
Theoretical Yield: The maximum amount of product that can be formed from a given amount of reactant, calculated based on the balanced equation of the reaction.
Actual Yield: The amount of product actually produced in a reaction.

Identify Given Information:
- Reactant quantity (e.g., methane or propane).
- Amount of product produced (actual yield).
- Combustion type of reaction.
Write the Balanced Chemical Equation:
- Recognize the hydrocarbon (e.g., methane CH4, propane C3H8) involved in the combustion reaction.
- Write the products: typically carbon dioxide and water.
Balancing the Chemical Equation:
- Count the number of each type of atom on both sides (reactants vs. products).
- Adjust coefficients to balance the equation, starting with carbon, then hydrogen, and finally oxygen.
Convert Amount to Grams or Moles:
- If the reactant is given in kilograms, convert to grams first (1 kg = 1000 g).
- Use the molar mass to convert grams to moles for stoichiometric calculations.
Using Stoichiometric Ratios:
- Determine the theoretical yield by converting moles of reactants to moles of products using the ratios from the balanced equation.
- Convert back to grams if needed.
Calculate Percent Yield:
- Use the formula:
[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 ]

Given 0.374 kg of methane, assess production of carbon dioxide (CO2).
Balanced Equation:
- CH4 + 2 O2 → CO2 + 2 H2O
- Balancing shows 1 C in reactants and products, 4 H in reactants and products, then balance O last.

Convert kg to g:
- 0.374 kg = 374 g.
Find Molar Mass of CH4:
- C: 12.01 g/mol, H: 1.01 g/mol (4 H)
- Molar Mass = 12.01 + (4 x 1.01) = 16.05 g/mol.
Convert grams to moles:
- Moles of CH4 = 374 g / 16.05 g/mol = 23.3 moles.

Stoichiometric Ratio: 1 mol CH4 produces 1 mol CO2.
Moles of CO2 = 23.3 moles of CH4 ➔ 23.3 moles of CO2.
Calculate grams of CO2:
- Molar Mass of CO2 = 12.01 + (2 x 16) = 44.01 g/mol.
- Grams of CO2 = 23.3 moles x 44.01 g/mol = 1025.70 g.

If actual yield of CO2 produced is given (e.g. 900 g), calculate percent yield:
- Convert grams of CO2 to same units and apply formula:
[ \text{Percent Yield} = \left( \frac{900 \text{ g}}{1025.70 \text{ g}} \right) \times 100 \approx 87.7% ]

Convert kg to g, find molar mass of propane (
- C: 12.01 (3 times) + H: 1.01 (8 times) = 44.11 g/mol).
Perform conversions and interrelations in moles.
Determine theoretical yield similarly through complete stoichiometric calculations.

Practice with different hydrocarbons for combustion analysis.
Remember significant figures during calculations for accurate results.
Typical question type for exams; ensure clarity on stoichiometry and yield calculations.

Identify Given Information:
- Reactant quantity (e.g., methane or propane).
- Actual yield (amount of product produced).
Write the Balanced Chemical Equation:
- Identify the hydrocarbon involved in the combustion reaction.
- Write down the products (typically carbon dioxide and water).
Balance the Chemical Equation:
- Count atoms on both sides (reactants vs. products).
- Adjust coefficients: balance carbon first, then hydrogen, and finally oxygen.
Convert Amount to Grams or Moles:
- If the reactant is in kg, convert to grams (1 kg = 1000 g).
- Use molar mass to convert grams to moles for calculation.
Use Stoichiometric Ratios:
- Convert moles of reactants to moles of products using the balanced equation ratios.
- Convert to grams if necessary.
Calculate Percent Yield:
- Use the formula: [ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 ]
Example Calculations:
- Follow through examples like combustion of methane or propane for practice.