Moles of Hydrofluoric Acid Produced from Xenon Compound Hydrolysis

To determine the number of moles of $\text{HF}$ produced upon complete hydrolysis of $1\text{ mol}$ of compound $\mathbf{P}$, we follow these steps:

Step 1: Identification of Compound P

Xenon ($\text{Xe}$) reacts with dioxygen difluoride ($\text{O}_2\text{F}_2$) at low temperature ($118\text{ K}$) to form xenon difluoride ($\text{XeF}_2$) and oxygen gas: $\text{Xe} + \text{O}_2\text{F}_2 \xrightarrow{118\text{ K}} \text{XeF}_2 + \text{O}_2$

Thus, the xenon compound $\mathbf{P}$ is $\text{XeF}_2$.

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Step 2: Hydrolysis of Compound P ($\text{XeF}_2$)

The complete hydrolysis of xenon difluoride ($\text{XeF}_2$) takes place according to the following balanced chemical equation: $2\text{XeF}_2\text{(s)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{Xe(g)} + 4\text{HF(aq)} + \text{O}_2\text{(g)}$

Dividing the coefficients by $2$ to represent the hydrolysis of $1\text{ mol}$ of $\text{XeF}_2$: $\text{XeF}_2 + \text{H}_2\text{O} \rightarrow \text{Xe} + 2\text{HF} + \frac{1}{2}\text{O}_2$

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Step 3: Calculation of Moles of HF

From the stoichiometry of the hydrolysis reaction: $\text{Moles of HF produced} = 2 \times (\text{Moles of } \text{XeF}_2)$

For $1\text{ mol}$ of compound $\mathbf{P}$ ($\text{XeF}_2$): $\text{Moles of HF produced} = 2 \times 1 = 2$

Final Answer: The number of moles of $\text{HF}$ produced is $2$.