# The complete combustion of 1 mole of pentane, C_5H_12, will require how many moles of oxygen, O_2?

Aug 20, 2017

$8$ ${\text{mol O}}_{2}$

#### Explanation:

We're asked to find the number of moles of ${\text{O}}_{2}$ needed to completely combust $1$ ${\text{mol C"_5"H}}_{12}$.

To do this, we first need to write the chemical equation for this reaction, following the general formula for a hydrocarbon combustion reaction:

$\text{hydrocarbon"color(white)(l) + "oxygen" rarr "carbon dioxide" + "water (vapor)}$

So

$\text{C"_5"H"_12(l) + "O"_2(g) rarr "CO"_2(g) + "H"_2"O"(g)" }$ (unbalanced)

Let's now balance this equation. We'll first put a "$5$" in front of the ${\text{CO}}_{2}$ to balance the carbons:

$\text{C"_5"H"_12(l) + "O"_2(g) rarr 5"CO"_2(g) + "H"_2"O"(g)" }$ (unbalanced)

Now we'll put a "$6$" in front of $\text{H"_2"O}$ to balance the hydrogens:

$\text{C"_5"H"_12(l) + "O"_2(g) rarr 5"CO"_2(g) + 6"H"_2"O"(g)" }$ (unbalanced)

We now have $2$ oxygens on the left side and $16$ on the right side. Putting an "$8$" in front of ${\text{O}}_{2}$ balances the equation:

ul("C"_5"H"_12(l) + 8"O"_2(g) rarr 5"CO"_2(g) + 6"H"_2"O"(g)

Now, we can realize that the stoichiometric ratio of pentane to oxygen is $1 : 8$ (from coefficients).

Therefore, in order to completely combust $1$ $\text{mol pentane}$, we must use color(red)(ul(8color(white)(l)"mol O"_2.