# combustion equation

## More: forum on burning

We start from the generic formula of complete combustion of alkanes:

CnH(2n+2) + (3n+1)/2*(O2+3.76N2) –> nCO2 + (n+1)H2O+(3n+1)/2*3.76N2

1) Study volume of complete combustion equation:

Considering the exhaust gas CNPT.
1 25 mole gas = L
Reason about the combustion of one mole of CnH fuel (+ 2n 2)

The above equation gives us exhaust:
25n L CO2
25 (n + 1) L H2O
25 (3n 1 +) / 2 3.76 * L N2

A total of 25n 25 + (n + 1) + 25 (3n 1 +) / * 2 3.76 25 = (7.64n 2.88 +) = n + 191 72 L gas.

Note: For n = 0 72 the L match the mole H2O and 1.88 mole N2 from the combustion of pure hydrogen.

For a given alkane so we have respectively:

25n / (191n 72 +)% of CO2
25 (n + 1) / (191n 72 +)% of H2O
(25(3n+1)/2*3.76)/(191n+72) % de N2

A division by 25 simplify formulas.

This applies in the case of complete combustion (no creation of CO or particulate) and ideal (no creation of NOx)

2) mass Study of the complete combustion equation:

Let us study the mass releases of the complete equation.

[CO2]=12+2*16=44 g/mol
[H2O] = 2 1 * + = 16 18 g / mol
[N2] = 2 14 * = 28g / mol

The calculation on the N2 is unnecessary in the case of an ideal combustion (no creation of Nox) because this element is not involved, it is an inert gas.

therefore the respective masses would be:
for CO2: 44n
for H2O: 18 (n + 1)

Application gasoline (octane pure). n = 8
[C8H18] = 8 12 * + = 18 1 114 * g / mol.
Mass CO2 rejected by mole octane consumed is: * 44 8 352 = g.
Mass H2O rejected by mole of octane is consumed: 18 (8 1 +) = g 162.
The fuel consumption report on releases of CO2 is 352 / 114 3.09 =

As the volume of the unit is more usual when speaking of fuel, it is better to spend this report CO2 gram per liter of gasoline consumed.

Knowing that the density of gasoline is 0.74 kg / l and 1 gram of gasoline burned rejected 3.09 grams CO2, we get: * 0.74 3.09 2.28 = CO2 kg per liter of gasoline burned.

These 2.28 kg occupy a volume of 2280 / 44 * = 25 1295 L CO2 released per liter of fuel consumed.

Similarly for H2O: The fuel consumption report on releases of CO2 is 162 / 114 1.42 =
where: * 0.74 1.42 1.05 = H2O kg per liter of gasoline burned.

Conclusion

A vehicle consuming 1 L of gasoline is going to reject a bit more than a kilo of water and 2.3 kg CO2.

The water will condense rather quickly directly or in the form of cloud and fall back into liquid form quite quickly (because we must not forget that water vapor is a very good greenhouse gas, much more "powerful" than the CO2), it is not the CO2 which has a life of the order of 100 years.

For other fuels, simply replace n by the fuel used. For example diesel consists of alkanes with n varying between 12 and 22. It would also intéréssant calculate releases of CO2 compared to the energy supplied by a given fuel. This will be the subject of another page.

Nevertheless, an article will follow the study of incomplete combustion (creating CO) and non-ideal (creating NOx)

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