# combustion equation

## More: forum on combustion

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) Volume study of the complete combustion equation:

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

The previous equation therefore gives us the escape:
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 the 72 L correspond to the mole of H2O and to the 1.88 mole of N2 resulting 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 is valid in the case of complete combustion (no creation of CO or particles) and ideal (no creation of Nox)

2) Mass study of the complete combustion equation:

Let us study the mass rejections 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 N2 is useless in the case of an ideal combustion (no creation of Nox) since this element does not intervene, it is an inert gas.

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

Application to petrol (pure octane). n = 8
[C8H18] = 8 12 * + = 18 1 114 * g / mol.
The mass of CO2 released per mole of octane consumed is: 44 * 8 = 352 g.
The mass of H2O released per mole of octane consumed is: 18 (8 + 1) = 162 g.
The ratio of fuel consumption to CO2 emissions is 352/114 = 3.09

As the unit of volume is more common when talking about fuel, it is preferable to pass this ratio in grams of CO2 per liter of petrol consumed.

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

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

Likewise for H2O: The ratio of fuel consumption to CO2 emissions is 162/114 = 1.42
hence: 0.74 * 1.42 = 1.05 kg of H2O per liter of fuel burned.

Conclusion

A vehicle consuming 1 L of petrol will therefore release a little more than a kilo of water and 2.3 kg of CO2.

Water will condense fairly quickly directly or in the form of a cloud and will fall back in liquid form fairly quickly (because we must not forget that water vapor is a very good greenhouse gas, much more "powerful" than CO2), this is not the case with CO2, which has a lifespan of around 100 years.

Read also: Diamonds are forever ... and against a hydraulic press?

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

Anyway, an article will follow with the study of incomplete combustion (creation of CO) and non-ideal (creation of Nox)