You're dealing with the combustion of propane. The balanced chemical equation looks like this
Now, the tool you have at your disposal when doing limiting reagent problems (or any stoichiometry problem) is the mole ratio. The balanced chemical equation gives you the proportion in which the reactans must mix; in this case, 1 mole of propane needs 5 moles of oxygen.
Calculate how many moles of each reactant you get by using their molar masses
The number of moles of oxygen you have doesn't even come close to this. This means that oxygen will determine how much propane reacts
To determine the mass of carbon dioxide produced use the mole ratio
Since you only used
The mass of excess propane will be
SIDE NOTE The mass of oxygen is so small that the limiting reagent was bound to be oxygen; the problem would have been a little more interesting (but not much) if you had 34.4 g of oxygen instead of 3.44...