Answer to Question #325304 in General Chemistry for Becky

Question

A solid hydrocarbon is burned in air in a closed container,producing a mixture of gases having a total pressure of 3.34 atm. Analysis of the mixture shows it to contain 0.340g of water vapour,0.792g of carbon dioxide,0.288g of oxygen,3.790g on nitrogen and no other gases. Calculate the mole fraction and partial pressure of carbondioxide in this mixture

Answer 1

χ_i = n_i / n

χ_i -  mole fraction of any individual gas component in a gas mixture

n - total moles of the gas mixture

n_i - moles of any individual gas component in a gas mixture

n = n(H₂O) + n(CO₂) + n(O₂) + n(N₂)

n(H₂O) = m(H₂O)/M(H₂O) = 0.340/18 = 0.019 moll

n(CO₂) = m(CO₂)/M(CO₂) = 0.792/44 = 0.018 moll

n(O₂) = m(O₂)/M(O₂) = 0.288/32 = 0.009 moll

n(N₂) = m(N₂)/M(N₂) = 3.790/28 = 0.135 moll

n = 0.019 moll + 0.018 moll + 0.009 moll + 0.135 moll = 0.181 moll

χ (CO₂) = n(CO₂)/n

χ (CO₂) = 0.018 moll / 0.181 moll = 0.099

p_i = χ_{i *} p

χ_i -  mole fraction of any individual gas component in a gas mixture

p_{i -} partial pressure of any individual gas component in a gas mixture

p - total pressure of the gas mixture

p(CO₂) = χ (CO₂) * p

χ (CO₂) = 0.099 * 3.34 = 0.331 atm