This page deals briefly with the combustion of alkanes and cycloalkanes. In fact, there is very little difference between the two.
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The word equation is: decane (a hydrocarbon) is transformed in water and carbon dioxide.
Complete combustion
Complete combustion (given sufficient oxygen) of any hydrocarbon produces carbon dioxide and water. It is quite important that you can write properly balanced equations for these reactions, because they often come up as a part of thermochemistry calculations. Some are easier than others. For example, with alkanes, the ones with an even number of carbon atoms are marginally harder than those with an odd number!
Example 1: Propane Combustion
For example, with propane (C3H8), you can balance the carbons and hydrogens as you write the equation down. Your first draft would be:
[ C_3H_8 + O_2 rightarrow 3CO_2 + 4H_2O]
Counting the oxygens leads directly to the final version:
[ C_3H_8 + 5O_2 rightarrow 3CO_2 + 4H_2O]
Example 2: Butane Combustion
With butane (C4H10), you can again balance the carbons and hydrogens as you write the equation down.
[ C_4H_{10} + O_2 rightarrow 4CO_2 + 5H_2O]
Counting the oxygens leads to a slight problem - with 13 on the right-hand side. The simple trick is to allow yourself to have 'six-and-a-half' O2 molecules on the left.
[ C_4H_{10} + 6dfrac{1}{2}, O_2 rightarrow 4CO_2 + 5H_2O]
If that offends you, double everything:
[ 2C_4H_{10} + 13 O_2 rightarrow 8CO_2 + 10 H_2O]
The hydrocarbons become harder to ignite as the molecules get bigger. This is because the bigger molecules don't vaporize so easily - the reaction is much better if the oxygen and the hydrocarbon are well mixed as gases. If the liquid is not very volatile, only those molecules on the surface can react with the oxygen. Bigger molecules have greater Van der Waals attractions which makes it more difficult for them to break away from their neighbors and turn to a gas.
Provided the combustion is complete, all the hydrocarbons will burn with a blue flame. However, combustion tends to be less complete as the number of carbon atoms in the molecules rises. That means that the bigger the hydrocarbon, the more likely you are to get a yellow, smoky flame.
Incomplete combustion![]()
Incomplete combustion (where there is not enough oxygen present) can lead to the formation of carbon or carbon monoxide. As a simple way of thinking about it, the hydrogen in the hydrocarbon gets the first chance at the oxygen, and the carbon gets whatever is left over! The presence of glowing carbon particles in a flame turns it yellow, and black carbon is often visible in the smoke. Carbon monoxide is produced as a colorless poisonous gas.
Why carbon monoxide is poisonous: Oxygen is carried around the blood by hemoglobin, which unfortunately binds to exactly the same site on the hemoglobin that oxygen does. The difference is that carbon monoxide binds irreversibly (or very strongly) - making that particular molecule of hemoglobin useless for carrying oxygen. If you breath in enough carbon monoxide you will die from a sort of internal suffocation.
1 Answer
Aug 18, 2017
Explanation:
The starting molecule in the question, undecane, will break down into three products. Two are listed in your question: ethene and propene (these are alkenes). The third product is an alkane that consists of the 'leftover' carbons and hydrogens. Ethene has the formula #C_2H_4#. Propene has the formula #C_3H_6#.
So far, you have:
#C_11H_24 -> C_2H_4 + C_3H_6 + C_?H_?#
Remember that we have to balance the equation, so the leftover carbons and hydrogens will go to the third product. We have #11-2-3 = 6# carbons and #24-4-6=14# hydrogens. This forms the alkane known as hexane.
Now the balanced equation is:
#C_11H_24 -> C_2H_4 + C_3H_6 + C_6H_14#
Cracking involves larger alkanes, so hexane should not break down.
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