Chemistry gas calculations

Question

Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0-L container. At 23.0 ∘C, the total pressure in the container is 5.30 atm . Calculate the partial pressure of each gas in the container.

find Pmethane, Pethane, Ppropane (atm)

Answer 1

Dalton's Law

Raoult's Law

moles CH₄ (methane) = 8.00 g CH₄ x 1 mol CH₄ / 16.0 g = 0.500 mols

moles C₂H₆ (ethane) = 18.0 g C₂H₆ x 1 mol C₂H₆ / 30.1 g = 0.598 mols

moles C₃H₈ (propane) = 1.08 mols (see calculations below)

Ideal gas law: PV = nRT

Solve for n (moles of gas)

n = PV / RT = (5.30 atm)(10.0 L) / (0.0821 Latm/Kmol)(296K)

n = 2.18 moles of gas

moles C₃H₈ = 2.18 mols - 0.500 mols - 0.598 mols = 1.08 mols C₃H₈

Partial pressure is determined by mole fraction of a gas x total pressure: X_gas * P_total = P_gas

X_CH4 = 0.500 mols / 2.18 mols = 0.229

X_C2H6 = 0.598 mols / 2.18 mols = 0.274

X_C3H8 = 1.08 mols / 2.18 mols = 0.495

X_gas * P_total = P_gas

P_CH4 = 0.229 x 5.30 atm = 1.21 atm

P_C2H6 = 0.274 x 5.30 atm = 1.45 atm

P_C3H8 = 0.495 x 5.30 atm = 2.62 atm