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

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#.