A 2.00 L reaction container at 25.0 °C contains CO2 (g) and O2 (g). The total pressure of the gases in the container was 656 torr and the mole fraction of CO2 is 0.245. A)Calculate the partial pressure of oxygen gas in the mixture.B)Calculate the mass of oxygen gas in the mixture (in g).C)Which gas will rise to the top of the container? Explain your answer.

The partial pressure (p) of a gas is related to the total pressure (Pt) of the gas by the mole fraction (X) as follows: p=Pt X ...(1). The partial pressure of the gas system is related to the moles of the gas (n) and temperature (T) as follows: n=pVRT ...(2). The density of the gas can be related to its raising ability in the atmosphere. The higher the density of the gas, the lower will be the height of the gas in the container. Determine the mole fraction of the oxygen and substitute in equation 1 along with the total pressure to calculate the partial pressure of oxygen gas. XO2=1-0.245=0.755PT=656 torrpO2=656 torr ×0.755=495.2 torrSubstitute the partial pressure of the oxygen, temperature (T), and the volume of the container (V) to calculate the number of moles of oxygen gas (n). nO2=pO2 VRTpO2=495.2 torr=0.001311 torr×495.2 torr=0.651 atmT=25.0°C=25.0°C+273=298 KV=2.00 LnO2=0.651 atm×2.00 L0.0820 L atm/K mol×298 K=0.0532 molmO2=n ×Molar mass=0.0532×32 g/mol=1.70 gThe density of carbon dioxide is much greater (1.977 g/L) than the density of oxygen (1.429 g/L). Hence in a given volume of the container, the carbon dioxide will stay at a lower height of the container and the oxygen moves up the container.