# How many moles of oxygen gas are required for the combustion of one mole of 3-methylpent-2-ene?

Oct 22, 2015

$\text{9 moles}$

#### Explanation:

The first thing to do here is determine the chemical formula of 3-methylpent-2-ene.

You know that you're dealing with an alkene, hence the suffix -ene.

Moreover, you know that the parent chain is pentane, and that the double bond is placed on the second carbon. This means that you will have a methyl group, $- {\text{CH}}_{3}$, placed on the third carbon, hence the prefix 3-methyl.

The compound will look like this - I'll show you (E)-3-methylpent-2-ene, which is the (E) geometric isomer of 3-methylpent-2-ene Notice that the compound's condensed formula can be written as ${\text{CH"_3"CHC"("CH"_3)"CH"_2"CH}}_{3}$.

This means that you can represent the compound as ${\text{C"_6"H}}_{12}$, since for its combustion reaction you only need to know how many carbon and how many hydrogen atoms it contains.

You're dealing with a hydrocarbon, which means that the combustion reaction will only produce carbon dioxide, ${\text{CO}}_{2}$, and water, $\text{H"_2"O}$.

So, the balanced chemical equation for the combustion of 3-methylpent-2-ene looks like this

$\text{C"_6"H"_12 + color(red)(9)"O"_2 -> 6"CO"_2 + 6"H"_2"O}$

Notice that you have a $1 : \textcolor{red}{9}$ mole ratio between 3-methylpent-2-ene and oxygen. This means that, in order for the reaction to take place, every mole of the former will require $\textcolor{red}{9}$ moles of the latter.

Therefore, one mole of 3-methylpent-2-ene will require nine moles of oxygen in order to react completely.