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5 December, 19:48

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.

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  1. 5 December, 23:26
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    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

    Partial Pressure of Methane = (0.499/2.263) (5.5 atm) = 1.21 atm

    Partial Pressure of Ethane = (0.599/2.263) (5.5 atm) = 1.46 atm

    Partial Pressure of Propane = (1.165/2.263) (5.5 atm) = 2.83 atm
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