(a) (2) Restatement: Find partial pressure of each gas.

Ptotal= Pfluorine+ Pxenon

.

..

P nRTV

80 10

4 868 0 0821 273

mole K liters

moles liter atm K

flourine 1

xenon

::#

:: :::

= = =1.4 atm

.

..

P nRTV

80 10

4 0 0 0821 273

mole K liters

moles liter atm K

xenon 1

xenon

::#

:: :::

= = =1.1 atm

(b) Restatement: 23.00 g Li added to flask. Weight of LiF formed?

Xe is inert (no reaction with lithium).

The balanced equation for the reaction is thus F 2 (g) + 2Li(s) →2 LiF(s).

.

1. .Li

23 00

694

gLi (^1331)
gLi

mole Li= mole

Therefore, lithium is the limiting reagent. So 3.31 moles of Li is used up along with
3.31/2 or 1.66 moles of F 2. This leaves 3.21 moles (4.868 −1.66) of F 2.

..

331

2

2

1

25 94

860

moles Li

moles Li

moles LiF

mole LiF

g LiF

##= g LiF

(c) Restatement: Partial pressures of gases present after reaction, 0°C

Xenon is inert (no reaction): 1.1 atm pressure

.

..

P nRTV

80 10

3 21 0 0821 273

mole K liters

moles liter atm K

fluorine 1

fluorine::

::#

:::

== = 0.90 atm

Part II: Specific Topics