Hi Brittany,
Yes, I agree that getting to see that combustion reaction under “invalid equations” would have helped in answering this question. However, recall from high school chemistry and/or first/second year intro to university chem class that combustion of hydrocarbons usually involve the hydrocarbon and O2 as reactants, then CO2 and H2O as products. So, pretty much the combustion of C12H23 would have this balanced equation:
4 C12H23 + 71 O2 —> 48 CO2 + 46 H2O
How I approached this is I divided both reactant and product side by 4, as we want to determine the number of moles of gas are produced (CO2 and H2O gas) from one mole of C12H23. In this case, we’ll get 12 mol CO2 and 11.5 mol H2O, which add to 23.5 moles. We then plug this value into PV=nRT to get V. In this case, our answer should be approximately 3400 L (Option D)
Hope these help!