Question

The reaction of $Xe$ and $O _2 F _2$ gives a Xe compound $P$. The number of moles of HF produced by the complete hydrolysis of $1 \,mol$ of $P$ is ______

Answer 1

Correct Answer: 4

To solve this problem, we need to analyze the reaction and the hydrolysis of the given compound to determine the number of moles of HF produced. 

1. Analyzing the Reaction:
From the reaction between \( \text{Xe} \) and \( \text{O}_2\text{F}_2 \), a xenon compound \( P \) is formed. The xenon compound formed is likely to be xenon tetrafluoride (XeF₄), as this is a common product in such reactions involving xenon and oxygen difluoride. The reaction can be represented as follows:

\[ \text{Xe} + \text{O}_2\text{F}_2 \rightarrow \text{XeF}_4 + \text{O}_2 \]

2. Hydrolysis of Xenon Tetrafluoride (XeF₄):
When xenon tetrafluoride (\( \text{XeF}_4 \)) undergoes complete hydrolysis, it reacts with water to produce xenon dioxide (XeO₂) and hydrofluoric acid (HF). The reaction for the hydrolysis of 1 mole of xenon tetrafluoride is:

\[ \text{XeF}_4 + 4\text{H}_2\text{O} \rightarrow \text{XeO}_2 + 4\text{HF} \]

3. Conclusion:
For each mole of xenon tetrafluoride (\( \text{XeF}_4 \)) hydrolyzed, 4 moles of hydrofluoric acid (HF) are produced.

Final Answer:
The number of moles of HF produced by the complete hydrolysis of 1 mole of \( P \) (which is \( \text{XeF}_4 \)) is 4.

Answer 2

\(\text{Xe} + 2\text{O}_2 + \text{F}_2 \rightarrow \text{XeF}_4 + 2\text{O}_2\)
\(3\text{XeF}_4 + 6\text{H}_2\text{O} \rightarrow 2\text{Xe} + \text{XeO}_3 + \frac{23}{2}\text{O}_2 + 12\text{HF}\)
∴ One mole of \(\text{XeF}_4\)​ gives 4 moles of HF on hydrolysis.