# What is the chemical equation for the combustion of C_3H_6?

Dec 14, 2015

$2 {\text{C"_3"H"_text(6(g]) + 9"O"_text(2(g]) -> 6"CO"_text(2(g]) + 6"H"_2"O}}_{\textrm{\left(l\right]}}$

#### Explanation:

Propene, ${\text{C"_3"H}}_{6}$, is a hydrocarbon, which is the term used to designate a compound that consists of only carbon and hydrogen.

When a hydrocarbon undergoes complete combustion, only two products are formed

• carbon dioxide, ${\text{CO}}_{2}$
• water, $\text{H"_2"O}$

This tells you that all the carbon that was a part of the hydrocarbon will now be a part of the carbon dioxide and all the hydrogen that was a part of the hydrocarbon will now be a part of water.

So, you know that one mole of propene contains

• three moles of carbon
• six moles of hydrogen

This tells you that the carbon dioxide produced by the reaction must contain three moles of carbon and the water must contain six moles of hydrogen.

Since one mole of carbon dioxide contains one mole of carbon, you can conclude that you need to have three moles of ${\text{CO}}_{2}$ in the products.

Likewise, since one mole of water contains two moles of hydrogen, you can say that you need to have three moles of water in the products.

Therefore, you can write

${\text{C"_3"H"_text(6(g]) + "O"_text(2(g]) -> 3"CO"_text(2(g]) + 3"H"_2"O}}_{\textrm{\left(l\right]}}$

Now, notice that the oxygen atoms are not balanced. More specifically, you have $2$ oxygen atoms on the reactants' side and a total of 9# oxygen atoms on the products' side.

Here is where a fractional coefficient comes in handy. Multiply the oxygen molecule by $\frac{9}{2}$ to get

${\text{C"_3"H"_text(6(g]) + 9/2"O"_text(2(g]) -> 3"CO"_text(2(g]) + 3"H"_2"O}}_{\textrm{\left(l\right]}}$

Finally, to get rid of this fractional coefficient, multiply all the compounds by $2$ to get the balanced

chemical equation for the combustion of propene

$2 {\text{C"_3"H"_text(6(g]) + 9"O"_text(2(g]) -> 6"CO"_text(2(g]) + 6"H"_2"O}}_{\textrm{\left(l\right]}}$