Problem: A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen was added to the bulb until the pressure was 72 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH, the final pressure of xenon and unreacted hydrogen in the bulb was 36 torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.)

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FREE Expert Solution

Balanced reaction:  XeF_x + (x/2) H₂ → Xe + xHF

According to Dalton’s Law:

$\boxed{{\mathbf{P}}_{\mathbf{total}}\mathbf{=}{\mathbf{P}}_{{\mathbf{XeF}}_{\mathbf{x}}}\mathbf{+}{\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}}}$

Before reaction: XeF_x + H₂

$$\begin{gathered} \mathbf{72}\mathbf{ }\mathbf{torr}\mathbf{=}\mathbf{18}\mathbf{ }\mathbf{torr}\mathbf{ }\mathbf{+}{\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}} \\ {\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}}\mathbf{=}\mathbf{72}\mathbf{ }\mathbf{torr}\mathbf{-}\mathbf{18}\mathbf{ }\mathbf{torr}\mathbf{ } \end{gathered}$$

PH₂ = 54 torr

• 1 mole XeF_x forms 1 mole Xe  → P_Xe = P_XeFx = 18 torr

Remaining in bulb after final reaction: Xe + unreacted H₂

• H₂ → excess reagent