What is the partial pressure of each gas in the mixture?

Question

A mixture of methane and oxygen gases, at a total pressure of 882 mm Hg, contains 1.45 grams of methane and 6.47 grams of oxygen.

Answer 1

The partial pressure of each gas will be equal to the mole fraction of each gas x the total pressure

Raoult's law

moles methane (CH₄) = 1.45 g x 1 mol/16 g = 0.0906 moles CH₄

moles oxygen (O₂) = 6.47 g x 1 mol/32 g = 0.202 moles O₂

Total moles of gas = 0.2928 moles

mol fraction CH₄ = 0.0906 mol/0.2928 mol = 0.309

mol fraction O₂ = 0.202 mol/0.2928 mol = 0.689

partial pressure CH₄ = mol fraction x total pressure = 0.309 x 882 mm Hg = 273 mm Hg

partial pressure O₂ = mol fraction x total pressure = 0.689 x 882 mm Hg = 608 mm Hg