Question

the partial pressure of CH4(g) is 0.175 atm and that of O2(g) is 0.250 atm in a mixture of the two gases.
A. what is the mole fraction of each gas in the mixture ?
B. If the mixture occupies a volume of 10.5 L at 65°C ,calculate the total number of moles of gas in the mixture
C. Calculate the number of grams of each gas in the mixture.
(answer in print not cursive)

Answer 1

(A.) Partial pressure CH₄ = 0.175 atm

Partial pressure O₂ = 0.250 atm

Total pressure = (Partial pressure CH₄) + (Partial pressure O₂)

Total pressure = (0.175 atm) + (0.250 atm)

Total pressure = 0.425 atm

mole fraction CH₄ = (Partial pressure CH₄) / (Total pressure)

mole fraction CH₄ = (0.175 atm) / (0.425 atm)

mole fraction CH₄ = 0.412

mole fraction O₂ = 1 - (mole fraction CH₄)

mole fraction O₂ = 1 - (0.412)

mole fraction O₂ = 0.588

(B.) Volume of mixture = 10.5 L

Temperature of mixture = 65^oC = (273 + 65) K = 338 K

According to ideal gas law,

Total moles = [(Total pressure) * (volume of mixture)] / [(R) * (absolute temperature of mixture)]

where R = gas constant = 0.0821 L-atm/mol-K

Total moles = [(0.425 atm) * (10.5 L)] / [(0.0821 L-atm/mol-K) * (338 K)]

Total moles = 0.161 mol

(C.) moles CH₄ in mixture = (mole fraction CH₄) * (total moles)

moles CH₄ in mixture = (0.412) * (0.161 mol)

moles CH₄ in mixture = 0.0663 mol

mass CH₄ in mixture = (moles CH₄ in mixture) * (molar mass CH₄)

mass CH₄ in mixture = (0.0663 mol) * (16.0 g/mol)

mass CH₄ in mixture = 1.06 g

moles O₂ in mixture = (mole fraction O₂) * (total moles)

moles O₂ in mixture = (0.588) * (0.161 mol)

moles O₂ in mixture = 0.0947 mol

mass O₂ in mixture = (moles O₂ in mixture) * (molar mass O₂)

mass O₂ in mixture = (0.0947 mol) * (32.0 g/mol)

mass O₂ in mixture = 3.03 g