Xenon fluoride can be prepared by heating a mixture of Xe and F2 gases to a high temperature in a pressure-proof container. Assume that xenon gas was added to a 0.25 liter container until its pressure reached 0.12 atm at 0.0°C. Fluorine gas was then added until the total pressure reached 0.72 atm at 0.0°C. After the reaction was complete, the xenon was consumed completely, and the pressure of the F2 remaining in the container was 0.36 atm at 0.0°C. What is the empirical formula of the xenon fluoride?

Dalton law's of partial pressure states that the total pressure equal to the sum of the partial pressure of the individual gases that make up the mixture.

Pt = Pxe + Pf₂

0.72 = 0.12 + Pf₂

0.72 - 0.12 = Pf₂

0.60 atm = Pf₂

after the the reaction was complete, the pressure of F₂ remaining = 0.36 atm

pressure of the consumed F₂ = 0.60 - 0.36 = 0.24 atm

Pxe / total pressure = number of mole / total number of mole of gas present

0.24 / 0.72 = nf / nt

0.12 / 0.72 = nxe / nt

comparing the two

(1/ 3) / ( 1/6) = (nf/ nt) / ( nxe/ nt)

nf / nxe = 2 / 1

the emperical formula = XeF₄

Sam1s Sam1s · Answer 2 · 2022-02-08T16:04:21+01:00

The empirical formula of the xenon fluoride is :

-XeF₄

"Dalton law's of partial pressure"

It states that the Total pressure break even with to the whole of the fractional weight of the person gasses that make up the blend.

Pt = Pxe + Pf₂
0.72 = 0.12 + Pf₂
0.72 - 0.12 = Pf₂
0.60 atm = Pf₂

Pressure of F₂ remaining = 0.36 atm

Pressure of the consumed F₂ = 0.60 - 0.36 = 0.24 atm

Pxe / total pressure = number of mole / total number of mole of gas present

0.24 / 0.72 = nf / nt

0.12 / 0.72 = nxe / nt

comparing the two:

(1/ 3) / ( 1/6) = (nf/ nt) / ( nxe/ nt)

nf / nxe = 2 / 1

The emperical formula = XeF₄

Learn more about "Emperical Formula" :

https://brainly.com/question/11588623?referrer=searchResults