Combustion Stoichiometry Calculator
Balance combustion equations for any hydrocarbon CnHm.
Calculates O2 required, CO2 and H2O produced, stoichiometric air-fuel ratio, and percent excess air.
Complete combustion of any hydrocarbon CnHm follows a predictable stoichiometry:
CnHm + (n + m/4) O2 -> n CO2 + (m/2) H2O
Carbon always becomes CO2; hydrogen always becomes water. The oxygen requirement changes with the fuel.
Methane (CH4, n=1, m=4): 1 + 1 = 2 moles O2. Produces 1 mole CO2 and 2 moles H2O. Octane (C8H18, n=8, m=18): 8 + 4.5 = 12.5 moles O2. Chemists multiply through by 2 to avoid fractions: 2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O. Propane (C3H8): 3 + 2 = 5 moles O2. C3H8 + 5 O2 -> 3 CO2 + 4 H2O.
Air is approximately 21% O2 by volume (the rest is mostly N2). So burning one mole of O2 requires 1/0.21 = 4.762 moles of air. This stoichiometric air-fuel ratio is critical in engine design: running rich (too little air) produces carbon monoxide; running lean (too much air) can cause knock and higher NOx emissions.
Excess air percentage = (actual air / stoichiometric air - 1) x 100%. Industrial combustion typically runs 10-20% excess air to ensure complete combustion while minimizing fuel waste. Power stations often target 3-5% excess O2 in flue gas as the control parameter.
Molecular weight of CnHm = 12n + m. The mass-based air-fuel ratio (AFR) = stoichiometric air moles x 28.97 / MW_fuel.
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