[blah2]

A 2L storage tank contains F₂(g) at 125 kPa, and 303K. A 1L reaction vessel contained Xe(g) at 50.4 kPa and 303K. The F₂ was pumped into the reaction vessel containing the Xe and the reaction heated for a few hours. Allowing the reaction Xe + 2F₂ ->XeF₄ to occur. After all the Xe had been used up, the reaction vessel cooled to 263K. Assuming all the XeF₄ remained in solid phase, what pressure of F₂ remained in the reaction vessel?


I could only get so far, I am not able to find the volume of F₂ that reacted.

Here is my attempt:
Xe + 2F₂ ->XeF₄
Since the two gases combine, the pressure is added.
Therefore P = 175.4 kPa
PV=nRT
n=PV/RT
n=(175.4kPa)(1L)/(8.31)(303K)
n=0.06966 mol Xe

Therefore 0.9966*2=molF₂
=0.13932mol F<sub>2

Don't know where to go from here.</sub>

[Borek]

No, you can't add pressures, that's completely off.

How many moles of Xe in reaction vessel before F₂ was added?

How many moles of F₂ in storage tank before it was transferred to reaction vessel?

[blah2]

Oh.
For F₂
PV=nRT
n=(PV)/(RT)
n=(125kPa)(2L)/(8.31)(303K)
n=0.09929 mol

For Xe
n=PV/RT
n=(50.4kPa)(1L)/(8.31)(303K)
n=0.02002 mol

[Borek]

OK, now it is a limiting reagent question. Do you see it?

[blah2]

I got it to the point where i found out the moles of F₂ that is unreacted, but for pv=nRT, i need a volume, so i am stuck there.

[Borek]

What is a volume of a reaction vessel?

[blah2]

oh 1 L, ty for your help