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 5.50 atm . calculate the partial pressure of each gas in the container.
1 answer:
Assume ideal gas behavior, then solve for the total number of moles:
PV = nRT
(5.50 atm)(10 L) = n(0.0821 L-atm/mol-K)(23+273 K)
n = 2.263 mol
Moles methane: 8 g ÷ 16.04 g/mol = 0.499 mol
Moles ethane: 18 g ÷ 30.07 g/mol = 0.599 mol
Moles propane: 2.263 - (0.499+0.599) = 1.165 mol
Applying Raoult's Law:
Partial pressure = Mole fraction * Total Pressure
<em>Partial Pressure of Methane = (0.499/2.263)(5.5 atm) = 1.21 atm</em>
<em>Partial Pressure of Ethane = (0.599/2.263)(5.5 atm) = 1.46 atm</em>
<em>Partial Pressure of Propane = (1.165/2.263)(5.5 atm) = 2.83 atm</em>
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