Question

The partial pressure of CH4(g) is 0.175 atm, and the partial pressure of O2(g) is 0.250 atm in a mixture of the two gases. The mixture occupies a volume of 10.5 L at 65 oC. Solve all three parts of the question. 1) The mole fraction of CH4(g) is , and the mole fraction of O2(g) is . 2) The total number of moles of gas in the mixture is . 3) There are grams of CH4(g) and grams of O2(g) in the mixture.

Answer 1

Answer:

1)

$x_{CH_4}=0.412\\x_{O_2}=0.588$

2)

$n_T=0.161mol$

3)

$m_{CH_4}=1.06gCH_4\\m_{O_2}=3.03gO_2$

Explanation:

Hello,

1) In this case, given the partial pressures of both methane and oxygen we can compute the mole fraction by considering each partial pressure as shown below:

$x_{CH_4}=\frac{P_{CH_4}}{P_T}=\frac{0.175atm}{0.175atm+0.250atm}=0.412\\x_{O_2}=\frac{P_{O_2}}{P_T}=\frac{0.250atm}{0.175atm+0.250atm}=0.588$

2) Now, we use the ideal gas equation to compute the total mass with the total pressure:

$PV=nRT\\\\n_T=\frac{PV}{RT}=\frac{(0.175+0.250)atm*10.5L}{0.082\frac{atm*L}{mol*K}*(65+273.15)K}=0.161mol$

3) Finally, by using the mole fractions, total moles and molar masses we compute the grams of both methane and oxygen:

$m_{CH_4}=0.412*0.161molCH_4*\frac{16gCH_4}{1molCH_4}=1.06gCH_4\\m_{O_2}=0.588*0.161molO_2*\frac{32gO_2}{1molO_2}=3.03gO_2$

Best regards.