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Stoichiometry of Combustion

Stoichiometry of Combustion refers to the examination of the relative proportions of fuel and oxygen required to carry out a complete combustion reaction.

When considering combustion reactions, chemists like to analyze its stoichiometric point, which is defined as the point where exactly all the oxygen is consumed and all the fuel has been burned. If there is too much oxygen compared to the amount of fuel present, then this is referred to as an overstoichiometric combustion. If there is not enough oxygen in the reaction, then this is referred to as an incomplete combustion – where not all the fuel has been burned.

Such a concept is important in the chemistry of engine function, when trying to determine the air-to-fuel ratio for maximum efficiency. For example, the stoichiometric point for methane would be calculated as follows:

CH4g) + 2O2(g) + 7.53N2(g) --> CO2(g) + 7.53N2(g) + 2H2O(l)

It is important to note that pure oxygen is rarely used in a combustion reaction. Instead pure air is used, which is comprised of 21% oxygen and 79% nitrogen. Converting them to the appropriate moles and coefficients for the stoichiometric point gives 2O2 (calculation: 2/2+7.53 = 21%) and 7.53N2 (calculation: 7.53/2+7.53 = 79%).

Therefore, it can be said that 1 mole of methane will react with 9.53 moles of air; and so the air-to-fuel ratio at the stoichiometric point is 9.53:1. Thus, understanding the stoichiometry of combustion is important for understanding the relative abundances of fuel and air in combustion reactions.

Combustion of methane, and mass of carbon dioxide, water and oxygen

When methane (CH4) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reaction is: CH4(g)+O2(g)>>CO2(g)+H2O(g) This type of reaction is referred to as a complete combustion reaction. What coefficients are needed to balance the equation for the complete combustion of

Energy, q, w, Enthalpy, and Stoichiometry

Consider the combustion of liquid methanol, CH3OH(l): CH3OH(l) + 3/2O2(g)>> CO2(g)+2H2O(l)ΔH = -726.5kJ 1)What is the enthalpy change for the reverse reaction? Express your answer using four significant figures. 2)Balance the forward reaction with whole-number coefficients: αCH3OH(l)+βO2(g)>>γCO2(g)+δH2O(l) G