In order to solve these kinds of problems, one must always use the balanced chemical equation and the ideal gas law, #PV = nRT#.

The balanced equation given is

#2C_4H_(10(g)) + 13O_(2(g)) -> 8CO_(2(g)) + 10H_2O((l))#

Notice that we have #2:8# (or a #1:4#) mole ratio between #C_4H_10# and #CO_2#; this means that for every mole of #C_4H_10# used in the reaction, **4** moles of #CO_2# will be produced.

Now, since we don't have a mass or a number of #C_4H_10# to go by, let's assume we start with #10.0g# of butane. Knowing that butane's molar mass is #58g/(mol)#, we can determine the number of moles from

#n_(butane) = m/(molarmass) = (10.0g)/(58g/(mol)) = 0.17# moles.

We thus get #n_(CO_2) = 4 * n_(bu t an e) = 0.68# moles.

So, the volume produced in this case is

#V = (nRT)/P = (0.68 * 0.082 * (273.15+23))/1.00 = 16.5L#

This method can be used for any mass of butane given...