A xenon fluoride can be prepared by heating a mixture of Xe and F 2 gases to a h

Question

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 0.75-L container until its pressure reached 0.60 atm at 0.0°C. Fluorine gas was then added until the total pressure reached 1.50 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 0.30 atm at 0.0°C. What is the empirical formula of the xenon fluoride gas?

Explanation / Answer

initially

PV = nRT

n = PV/(RT) = (0.60*0.75)/(0.082*273) = 0.0201 mol of Xe added

then..

addition :

PV = nRT

ntotal = PV/(RT) = (1.5*0.75)/(0.082*273) = 0.05025

mol of F2 = total - mol of Xe = 0.05025 -0.0201 = 0.03015 mol of F2

after reaction, all Xe is reacted, then

n = PV/(RT) = 0.3*0.75/(0.082*273) = 0.010

mol of F2 left = 0.010

mol of F2 reacted = 0.03015 -0.010 = 0.02015

mol of Xe reacted = 0.0201

ratio Xe:F2 = 0.02015/0.0201 = 1

that is

Xe + F2 = XeF2 forms

then, empirical formula is XeF2

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