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 = mole fraction x total pressure

toal pressure = 0.175+0.250 = 0.425 atm

mole fraction of CH₄ = partial pressure of CH₄ / total pressure = 0.175 / 0.425 = 0.412

mole fraction of O₂ = partial pressure of CH₄ / total pressure = 0.250 / 0.425 = 0.588

B. P= 0.425, V= 10.5 L , T= 273.15 +65 = 338.15 K, R = 0.0821 L. atm / mol. K

PV=nRT

n = PV/RT = (0.425 x 10.5 ) / (0.0821 x 338.15) = 4.4625 = 27.762 = 0.161 moles

C.

moles = mole fraction x total moles & mass = molar mass x moles molar mass of CH₄ = 16 g/mol & molar mass of O₂ = 16 g/mol

moles of CH₄ = 0.412 x 0.161 = 0.0663 mole

mass of CH₄ = 16 x 0.0663 = 1.061 g

moles of O₂ = 0.588 x 0.161 = 0.0947 mole

mass of O₂ = 32 x 0.0947 = 3.030 g