This is a pretty straightforward limiting reagent problem, with the added twist that you're working at a molecular level.
The balanced chemical equation for this combustion reaction looks like this
#"C"_ 4"H"_ (8(g)) + color(red)(6)"O"_ (2(g)) -> 4"CO"_ (2(g)) + 4"H"_ 2"O"_((l))#
Usually, you would look at this chemical equation and say that butene,
But as you know, a mole is simply a very, very large collection of molecules. This means that you an think about the mole ratio as being a molecule ratio.
In this case, you can say that
Now, your goal here is to determine which of the two reactants acts as a limiting reagent, i.e. which reactants will be completely consumed before all the molecules of the other reactant get the chance to react.
Let's assume that all the
#28 color(red)(cancel(color(black)("molecules C"_4"H"_8))) * (color(red)(6)color(white)(a)"molecules O"_2)/(1color(red)(cancel(color(black)("molecule C"_4"H"_8)))) = "168 molecules O"_2#
Since you have more molecules of oxygen gas available, oxygen gas will be in excess. In other words, butene will act as a limiting reagent.
These butene molecules will consume
#"remaining O"_2 = 228 - 168 = color(green)(|bar(ul(color(white)(a/a)"60 molecules O"_2color(white)(a/a)|)))#
Once the reaction is finished, you will be left with