Like any stoichiometry problem, if we want to compare quantities of different reactants or products, we must do so in moles. Therefore, we can start by finding the number of moles of O2 combusted by using the ideal gas law. Don’t forget to use Kelvin temperatures!
PV = nRT; (1 atm)(82.9 L) = (n)(0.0821 L*atm/K*mol)(273 K); n = 3.70 mol O2
With the number of moles of O2 in hand, we can first convert to moles of pentane using our mole-to-mole ratio, then to grams of pentane using its molar mass.
3.70 mol O2 * 1 mol C5H12/8 mol O2 * 72.151 g C5H12/1 mol C5H12 = 33.4 g C5H12
I hope this helped!