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.

Answer 1

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

Answer 2

The partial pressure for methane, Ethane and propane is 1.21 atm, 1.24 atm, 2.83 atm respectively.

Given here,

Mass of methane 8g

Mass of ethane is 18g

From ideal gas formula

PV = nRT

(5.50 atm)(10 L) = n(0.0821 )(296 K)

n = 2.263 mol

Moles methane = 0.499 mol

Moles ethane = 0.599 mol

Moles propane = 1.165 mol

According to Raoult's law

$\rm \bold { Pa = P' \times Xa}$

where,

Pa is partial pressure

P' is total pressure

Xa is mole fraction

By solving the equation for each

We can conclude that the partial pressure for methane, Ethane and propane is 1.21 atm, 1.24 atm, 2.83 atm respectively.

To know more about Raoult's Law, refer to the link:

https://brainly.com/question/17760155?referrer=searchResults