Combustion stoichiometry is the calculation of the exact reactant amounts needed for a fuel to burn completely. Engineers use it to size air supplies, estimate exhaust products, compare fuels, and evaluate engines, burners, boilers, and furnaces. This cheat sheet gives students a clean reference for balancing combustion reactions and converting between moles, mass, air, and products.
It is especially useful when solving multi-step problems where small mole-ratio errors can change the final answer.
Key Facts
- For complete combustion of a hydrocarbon CxHy, the balanced reaction is CxHy + (x + y/4) O2 -> x CO2 + (y/2) H2O.
- For complete combustion of CxHyOz, the theoretical oxygen requirement is x + y/4 - z/2 moles O2 per mole of fuel.
- Dry air is commonly modeled as 21 percent O2 and 79 percent N2 by mole, so 1 mole O2 is supplied with 3.76 moles N2.
- Theoretical air in moles equals required moles O2 divided by 0.21, or equivalently required moles O2 times 4.76.
- Stoichiometric air-fuel ratio by mass is AFRstoich = mass of theoretical air / mass of fuel.
- Percent excess air is percent excess air = (actual air - theoretical air) / theoretical air x 100 percent.
- Equivalence ratio is phi = (fuel-air ratio actual) / (fuel-air ratio stoich) = (AFRstoich) / (AFRactual).
- If phi < 1, the mixture is lean with excess oxygen, and if phi > 1, the mixture is rich with insufficient oxygen for complete combustion.
Vocabulary
- Stoichiometric combustion
- Combustion with exactly enough oxygen to convert all fuel carbon to CO2 and all fuel hydrogen to H2O with no excess oxygen.
- Theoretical air
- The minimum amount of air needed for complete combustion of a given amount of fuel.
- Excess air
- Air supplied above the theoretical air requirement, usually expressed as a percent of theoretical air.
- Air-fuel ratio
- The mass or mole ratio of air supplied to fuel supplied in a combustion process.
- Equivalence ratio
- A dimensionless measure of mixture richness defined as actual fuel-air ratio divided by stoichiometric fuel-air ratio.
- Flue gas
- The combustion product gas mixture leaving a burner, furnace, engine, or boiler.
Common Mistakes to Avoid
- Forgetting nitrogen from air, which is wrong because air is not pure oxygen and each mole of O2 brings about 3.76 moles of N2 into the products.
- Using mass ratios when the balanced equation gives mole ratios, which is wrong because coefficients in chemical equations represent moles, not kilograms or grams.
- Ignoring oxygen already in the fuel formula, which is wrong because a fuel such as C2H5OH needs less outside O2 than a hydrocarbon with the same carbon and hydrogen counts.
- Confusing percent excess air with percent oxygen in products, which is wrong because excess air compares actual air to theoretical air, not exhaust composition directly.
- Treating rich combustion as complete combustion with leftover oxygen, which is wrong because rich mixtures have too little oxygen and may form CO, unburned fuel, or soot.
Practice Questions
- 1 Balance the complete combustion equation for methane: CH4 + O2 -> CO2 + H2O, then include N2 using air as 21 percent O2 and 79 percent N2 by mole.
- 2 For 1 mole of propane, C3H8, calculate the theoretical moles of O2 and theoretical moles of air required for complete combustion.
- 3 A burner needs 10.0 kg of theoretical air but is supplied with 12.5 kg of actual air. Calculate the percent excess air.
- 4 Explain why an engine might be operated with excess air even though stoichiometric combustion uses the exact required amount of oxygen.