Question #cb618

1 Answer
Apr 5, 2016


#"60 molecules O"_2#


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, #"C"_4"H"_8#, reacts in a #1:color(red)(6)# mole ratio with oxygen gas, #"O"_2#.

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 #1# molecule of butene needs #color(red)(6)# molecules of oxygen gas in order for the combustion reaction to take place.

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# molecules of butene get the chance to react. This means that you will need

#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 #168# molecules of oxygen gas, which means that you'll be left with

#"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 #0# molecules of butene and #60# molecules of oxygen gas.