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

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 5.50 atm . calculate the partial pressure of each gas in the container.

Abbie Toy · Accepted 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 molMoles methane: 8 g ÷ 16.04 g/mol = 0.499 molMoles ethane: 18 g ÷ 30.07 g/mol = 0.599 molMoles propane: 2.263 - (0.499+0.599) = 1.165 molApplying Raoult's Law: Partial pressure = Mole fraction * Total PressurePartial Pressure of Methane = (0.499/2.263)(5.5 atm) = 1.21 atmPartial Pressure of Ethane = (0.599/2.263)(5.5 atm) = 1.46 atmPartial Pressure of Propane = (1.165/2.263)(5.5 atm) = 2.83 atm

Abbie Toy · Answer

Partial pressure of methane: 1.18 atmPartial pressure of ethane: 1.45 atmPartial pressure of propane: 2.35 atmExplanation: Let the total moles of gases in a container be n.Total pressure of the gases in a container =P = 5.0 atmTemperature of the gases in a container =T = 23°C = 296.15 KVolume of the container = V = 10.0 L (Ideal gas equation)Moles of methane gas =Moles of ethane gas =Moles of propane gas =Partial pressure of all the gases can be calculated by using Raoult's law: = partial pressure of 'i' component. = mole fraction of 'i' component in mixtureP = total pressure of the mixturePartial pressure of methane:Partial pressure of ethane:Partial pressure of propane:

Mr. Ellis Hintz

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

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