A mixture consisting of 0.120 mol N2, 0.018 mol O2, 0.112 mol CH4, and an unknown amount of CO2 occupies a volume of 9.04 L at 25 Solution V(N2) = nRT/P = 0.12x0.0821x298/1.17) = 2.509 l, V(O2) = 0.018x0.0821x298/1.17 =0.3764 L, V(CH4) = 0.112x0.0821x298/1.17 = 2.342 L, V(CO2) = 9.04-2.509-0.3764-2.342 = 3.81257, nCO2 = PV/RT = 1.17 x3.81257/(0.0821x298) = 0.1823 moles.