A closed container has a mixture of methane, CH4, ethane C2H6, and propane C3H8. If 12.5 g of methane, 36.0 g of ethane, and 14.0 g of propane have a total pressure of 890. mm Hg, what is the partial pressure of each of the gases in the container?
Convert grams each gas to mols.
mols = g/molar mass
Convert each gas to mole fraction.
Xeach gas = mol each gas/total mols.
partial pressure each gas = Xeach gas*Ptotal
To find the partial pressure of each gas in the container, we'll use Dalton's law of partial pressures. According to Dalton's law, the total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of each individual gas.
To find the partial pressures, we need to determine the moles of each gas using the given masses and molar masses. Then, we can calculate the partial pressure using the ideal gas law.
First, let's calculate the moles of each gas:
- Methane (CH4):
Moles = Mass / Molar Mass = 12.5 g / 16.04 g/mol = 0.779 mol
- Ethane (C2H6):
Moles = Mass / Molar Mass = 36.0 g / 30.07 g/mol = 1.198 mol
- Propane (C3H8):
Moles = Mass / Molar Mass = 14.0 g / 44.11 g/mol = 0.317 mol
Next, let's calculate the partial pressure of each gas:
- Methane (CH4):
PCH4 = (nCH4 / nTotal) × PTotal
PCH4 = (0.779 mol / (0.779 mol + 1.198 mol + 0.317 mol)) × 890. mm Hg
PCH4 = 0.322 × 890. mm Hg = 287 mm Hg (rounded to three significant figures)
- Ethane (C2H6):
PC2H6 = (nC2H6 / nTotal) × PTotal
PC2H6 = (1.198 mol / (0.779 mol + 1.198 mol + 0.317 mol)) × 890. mm Hg
PC2H6 = 0.489 × 890. mm Hg = 434 mm Hg (rounded to three significant figures)
- Propane (C3H8):
PC3H8 = (nC3H8 / nTotal) × PTotal
PC3H8 = (0.317 mol / (0.779 mol + 1.198 mol + 0.317 mol)) × 890. mm Hg
PC3H8 = 0.189 × 890. mm Hg = 169 mm Hg (rounded to three significant figures)
Therefore, the partial pressure of methane is approximately 287 mm Hg, the partial pressure of ethane is approximately 434 mm Hg, and the partial pressure of propane is approximately 169 mm Hg.