Stoichiometric Ratio: complete combustion of hydrocarbon

Stoichiometric Ratio: complete combustion of hydrocarbon

Complete combustion of a hydrocarbon fuel requires sufficient amount of air to convert fuel completely to carbon dioxide and water vapor. Since, 23% by mass of oxygen in air participate in combustion the stoichiometric air fuel ratio can be calculated from the combustion equation.

For example, one mole of C₇H₁₆ requires 11 moles of oxygen for complete combustion.

                             C₇H₁₆ + 11O₂ => 7CO₂ + 8H₂O

By substituting appropriate atomic weights

C = 12 H= 1 O=16

We get

100g + 352g = 308g + 144g

Thus, 1g of fuel requires 3.52g of oxygen or

3.52 x (100/23) = 15.3g of air.

Thus the stoichiometric Air-Fuel Ratio (AFR) for C₇H₁₆ is 15.3:1.

Stoichiometric mixture contains by definition sufficient amount of oxygen for complete combustion of fuel. Thus, at the stoichiometric AFR, the fuel will release all of its latent heat of combustion. If combustion is carried out at AFR higher than stoichiometric AFR, it is called lean or weak mixture. Combustion at an AFR lower than the stoichiometric value implies a deficiency of oxygen hence, combustion is incomplete and partially burned fuel principally in the form of carbon monoxide and unburned hydrocarbon will escape from the combustion zone. 


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