Combustion Reaction Calculator

CxHy + (x + y/4) O2 -> x CO2 + (y/2) H2O. For propane (C3H8): C3H8 + 5O2 -> 3CO2 + 4H2O. If the molecule contains oxygen (CxHyOz), subtract the oxygen already present: the O2 coefficient becomes (x + y/4 - z/2). Complete combustion requires excess oxygen; insufficient oxygen produces CO and soot (incomplete combustion).

Gasoline is approximately C8H18 (octane). Burning 1 gallon (~2.8 kg, about 24.5 mol of octane) produces 24.5 x 8 = 196 mol of CO2, which weighs 196 x 44.009 = 8,625 g or about 8.6 kg of CO2. The EPA figure of 8.89 kg CO2 per gallon of regular gasoline (which includes the full carbon content) is consistent with this estimate.

The stoichiometric air-to-fuel ratio is the exact mass of air needed for complete combustion. For gasoline it is about 14.7:1 (14.7 kg air per kg fuel). A lean mixture (> 14.7:1) has excess air - cleaner combustion but less power. A rich mixture (< 14.7:1) has insufficient air - incomplete combustion, more CO and unburned hydrocarbons. Modern engines use oxygen sensors to maintain the stoichiometric ratio.

Stoichiometric combustion supplies exactly the oxygen the balanced equation requires, with none left over. This calculator solves that ideal case. Real burners deliberately supply more than that so every fuel molecule finds oxygen, since mixing is never perfect. The extra is called excess air, and the ratio of actual to stoichiometric air is the equivalence-ratio lambda (λ): λ = 1 is stoichiometric, λ > 1 is lean (excess air), and λ < 1 is rich (excess fuel). For example, 20% excess air means λ = 1.2 and 1.2 times the theoretical air. Excess air ensures complete combustion and limits CO, but too much of it wastes heat up the flue and raises NOx, so industrial boilers typically run around 10 to 20% excess air (λ ≈ 1.1 to 1.2).