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

1 Answer
Aug 20, 2017

8 "mol O"_2

Explanation:

We're asked to find the number of moles of "O"_2 needed to completely combust 1 "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:

"hydrocarbon"color(white)(l) + "oxygen" rarr "carbon dioxide" + "water (vapor)"

So

"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 "CO"_2 to balance the carbons:

"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 "H"_2"O" to balance the hydrogens:

"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 "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 "mol pentane", we must use color(red)(ul(8color(white)(l)"mol O"_2.