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...
