Examples of Combustion Reaction

We explain that what are examples of combustion reaction? combustion is a chemical reaction in which an organic substance , is, mainly formed by carbon and hydrogen, decays by emitting heat, gases and water vapor . Its energy released in the form of heat appears as a flame. Depending on the complexity of the molecule, the combustion will release a quantity of heat. If it is more complex, it will be greater.

For the combustion reaction to occur, three factors are necessary:

  • The organic substance , which generally has to be in a liquid or gaseous state in order to rapidly react. Its role in combustion is like fuel . All organic substances are combustible.
  • spark or a high temperature that can lead the organic substance to ignition, that is, the initiation of combustion.
  • There must be an oxidizer , which is a chemical reagent that is going to combine with the carbon and hydrogen atoms that are released from the organic substance. It is therefore considered to support combustion. The oxygen contained in the air is the oxidizer par excellence.

The fuels

It is called fuel to any substance capable of being decomposed in a combustion reaction. These are organic compounds, which are made up mainly of carbon and hydrogen atoms. The hydrocarbons , which are alkanes, alkenes and alkynes, stand out as fuels.

The reaction of hydrocarbons is rapid to the spark and delivers sufficient heat that can be directed to increase the temperature of many substances and containers. The simplest hydrocarbon is methane (CH 4 ) . This alkane is the main component of natural gas , which is used in domestic and industrial stoves.

Higher alkanes, such as propane (C 3 H 8 ) and butane (C 4 H 10 ), form LP gas (liquefied petroleum gas), which is obtained from the distillation of petroleum. It is one of the most volatile components of the latter. The propane (C 3 H 8 ) is in a 70% and butane (C 4 H 10 ) in 30% . LP gas is a more powerful fuel than natural gas , since it contains more carbons.

Theoretical air and excess air

When combustion is to be carried out in industrial equipment, air is going to be supplied so that the oxygen that composes it reacts with the fuel and acts as an oxidizer. Oxygen (O 2 ) is found in the air in a percentage of 21% . That is why first, based on the amount of fuel, the amount of oxygen needed will be calculated.

Knowing the necessary oxygen (measured in moles), this is going to be translated into an amount of air to make it react with the fuel. This is called theoretical air , and it is obtained from the chemical equation of combustion. When the industrial plant adds a quantity greater than this, it will be called excess air , and it is not random, but a certain percentage of the theoretical one.

Example:

To calculate the theoretical air for a combustion reaction, we start from the chemical equation:

CH 4 + 2O 2 -> CO 2 + 2H 2 O

The equation is going to be interpreted first: 1 mole of CH 4 reacts with 2 moles of O 2 , to produce 1 mole of CO 2 and 2 moles of H 2 O. That is the stoichiometry or way of describing the relationship between them.

Now we start from a calculation base of 100 moles of methane CH 4 fuel . It would have to:

  • 100 moles of CH 4 react
  • 200 moles of O 2 are needed for the reaction
  • 100 moles of CO 2 are produced
  • 200 moles of H 2 O are produced

The moles of oxygen O must be translated into moles of air .

So, to know how many moles of air are going to be added to the combustion reaction, the following calculation is made:

2 is 21% of the air: there are 200 moles of O 2

Air is 100%: is there? Air moles

A rule of three is made: [(200 moles of O 2 ) * (100%)] / (21%) = 952.3809 moles of air

If you want to transform to grams , use the molecular weight of air.

The molecular weight of air is calculated, for practical purposes, as:

  • 21% with the molecular weight of molecular oxygen (O 2 )
  • 79% with the molecular weight of molecular nitrogen (N 2 )

The molecular weight of air is:

(0.21 * 32g / mol) + (0.79 * 28g / mol) = 6.72 + 22.12 = 28.84 g / mol

To transform into grams, the moles of air are multiplied by the molecular weight:

Grams of air = (Moles of air) * (Molecular weight of air)

Grams of air = (952.3809 moles of air) * (28.84 g / mol)

Grams of air = 27466.66 grams = 27.46 kilograms of air

The latter is the theoretical air, and the excess air can be added as a percentage of that.

Starting from the previous example, the theoretical air percentages are calculated:

Theoretical air = 27.46 kilograms of air

Air with 10% in excess = (27.46 Kg) * (1.10) = 30.206 Kg

Air with 20% in excess = (27.46 Kg) * (1.20) = 32.952 Kg

Air with 30% in excess = (27.46 Kg) * (1.30) = 35.698 Kg

Complete and incomplete combustion

It is said that combustion is complete when the reaction produces only carbon dioxide (CO 2 ) and water vapor (H 2 O). An incomplete combustion is, instead, when it produces carbon dioxide (CO 2 ), water vapor (H 2 O) and also carbon monoxide (CO). Ideally, the oxygen is combined with the fuel to produce only CO 2 .

When combustion is incomplete and only carbon monoxide is produced, the fuel is not being used efficiently and its heat cannot be used for industrial processes.

20 Examples of combustion reactions

1.- Combustion of methane (CH 4 )

CH 4 + 2 O 2 -> CO 2 + 2 H 2 O

2.- Combustion of ethane (C 2 H 6 )

2 H 6 + 7/2 O 2 -> 2 CO 2 + 3 H 2 O

3.- Combustion of propane (C 3 H 8 )

3 H 8 + 5 O 2 -> 3 CO 2 + 4 H 2 O

4.- Combustion of butane (C 4 H 10 )

4 H 10 + 13/2 O 2 -> 4 CO 2 + 5 H 2 O

5.- Combustion of pentane (C 5 H 12 )

5 H 12 + 8 O 2 -> 5 CO 2 + 6 H 2 O

6.- Combustion of hexane (C 6 H 14 )

6 H 14 + 19/2 O 2 -> 6 CO 2 + 7 H 2 O

7.- Combustion of heptane (C 7 H 16 )

7 H 16 + 11 O 2 -> 7 CO 2 + 8 H 2 O

8.- Combustion of octane (C 8 H 18 )

8 H 18 + 25/2 O 2 -> 8 CO 2 + 9 H 2 O

9.- Combustion of nonane (C 9 H 20 )

9 H 20 + 14 O 2 -> 9 CO 2 + 10 H 2 O

10.- Combustion of the dean (C 10 H 22 )

10 H 22 + 31/2 O 2 -> 10 CO 2 + 11 H 2 O

11.- Combustion of undecane (C 11 H 24 )

11 H 24 + 17 O 2 -> 11 CO 2 + 12 H 2 O

12.- Combustion of dodecane (C 12 H 26 )

12 H 26 + 37/2 O 2 -> 12 CO 2 + 13 H 2 O

13.- Combustion of tridecane (C 13 H 28 )

13 H 28 + 20 O 2 -> 13 CO 2 + 14 H 2 O

14.- Combustion of tetradecane (C 14 H 30 )

14 H 30 + 43/2 O 2 -> 14 CO 2 + 15 H 2 O

15.- Combustion of pentadecane (C 15 H 32 )

15 H 32 + 23 O 2 -> 15 CO 2 + 16 H 2 O

16.- Combustion of hexadecane (C 16 H 34 )

16 H 34 + 49/2 O 2 -> 16 CO 2 + 17 H 2 O

17.- Combustion of heptadecane (C 17 H 36 )

17 H 36 + 26 O 2 -> 17 CO 2 + 18 H 2 O

18.- Combustion of octadecane (C 18 H 38 )

18 H 38 + 55/2 O 2 -> 18 CO 2 + 19 H 2 O

19.- Combustion of nonadecane (C 19 H 40 )

19 H 40 + 29 O 2 -> 19 CO 2 + 20 H 2 O

20.- Combustion of eicosane (C 20 H 42 )

20 H 42 + 61/2 O 2 -> 20 CO 2 + 21 H 2 O

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