   # A xenon fluoride can be prepared by heating a mixture of Xe and F 2 gases to a high temperature in a pressure-proof container. Assume that xenon gas was added to a 0.25-L 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 F 2 remaining in the container was 0.36 atm at 0.0 °C. What is the empirical formula of the xenon fluoride? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 10, Problem 78GQ
Textbook Problem
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## A 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-L 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?

Interpretation Introduction

Interpretation:

The Xenon Fluoride compound prepared from (Xe) and (F2) gas, with high temperature and pressure.  In this statement both reactants pressure and temperature are given, using this values the (XeF2) empirical formula should identified.

### Explanation of Solution

Given data:

Pressure of xenon is 0.12atm.  Total pressure of gas in container is 0.72atm.  Temperature is 0°C.  Volume of the container is 0.25L.

Conversion of temperature to kelvin scale:

T(K)=(T(°C)+273.15K)=0+273.15K=273.15K

Pressure of fluorine gas can be calculated as shown below.

P=PXe+PF2PF2=PPXe=0.72atm0.12atm=0.60atm

Number of moles of xenon gas is calculated as shown below.

PV=nRTn=PVRT=0.12atm×0.25L0.08205LatmK1×273.15K=0.0322.4119575=1.34×10-3

Number of moles of fluorine gas is calculated as shown below.

PV=nRTn=PVRT=0.60atm×0.25L0.08205LatmK1×273

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