When chemical reactions occur, the bonds between molecules of reagents will be destroyed and atoms, electrons will be rearranged to form product. The energy will be formed when the rapid combustion process is performed by combustible elements. The fuel can be burnt completely if the carbon element contained in the fuel is burnt to be carbon dioxide or all of hydrogen element is burnt to be water and sulfur to be sulfur dioxide.
As an illustration of the theoretical amount of air in the combustion of methane, the combustion product of this reaction contains only carbon dioxide, water and nitrogen.
CH4 + a(O2 + 3.76 N2) → bCO2 + cH2O + dN2
The total mass of each chemical element must be same on both sides of the equation. Although there are different elements of their chemical compounds in the reagent and reaction, amount of moles of reagent with combustion product may differ. Amount of air to supplies sufficient oxygen to perform complete combustion of carbon, hydrogen and sulfur contained in the fuel is called stoichiometric amount of air.
The parameter of equation above are a, b, c, d represents the number of moles of oxygen, water, carbon dioxide and nitrogen. Value 3.76 moles of nitrogen are considered to accompany oxygen. The following value is taken by applying the mass conservative of carbon, hydrogen, oxygen and nitrogen:
C: b = 1
H: 2c = 4
O: 2b + c = 2a
N: d = 3.76 a
So the equation above will be:
CH4 + 2(O2 + 3.76 N2) → CO2 + 2H2O + 7.52N2
Coefficient 2 before (O2 + 3.76 N2) is the number of moles of oxygen in the combustion air per mole of fuel not the amount of air. The amount of combustion air is 2 mole oxygen plus 2 x 3.76 moles of nitrogen to give 9.52 mole air per mole of fuel.