The reaction of Xe and O₂F₂ gives a Xe compound P. The number of moles of HF produced by the complete hydrolysis of 1 mol of P is______.

The correct answer is 4.

Step 1: Reaction of Xe with O₂F₂
When xenon ($\text{Xe}$) reacts with dioxygen difluoride (${\text{O}}_2{\text{F}}_2$), xenon tetrafluoride (${\text{XeF}}_4$) is formed along with dioxygen gas.
The balanced chemical equation for this reaction is:

$\text{Xe}+2{\text{O}}_2{\text{F}}_2\to {\text{XeF}}_4+2{\text{O}}_2$

Thus, compound P is xenon tetrafluoride, ${\text{XeF}}_4$.

Step 2: Hydrolysis of compound P (XeF₄)
When xenon tetrafluoride (${\text{XeF}}_4$) undergoes complete hydrolysis, it disproportionates to yield xenon ($\text{Xe}$), xenon trioxide (${\text{XeO}}_3$), hydrofluoric acid ($\text{HF}$), and oxygen gas (${\text{O}}_2$).
The balanced chemical equation for the hydrolysis reaction is:

$6{\text{XeF}}_4+12{\text{H}}_2\text{O}\to 4\text{Xe}+2{\text{XeO}}_3+24\text{HF}+3{\text{O}}_2$

Step 3: Calculating the number of moles of HF produced per mole of P
Dividing the entire balanced equation by 6 to find the stoichiometry for 1 mole of ${\text{XeF}}_4$:

${\text{XeF}}_4+2{\text{H}}_2\text{O}\to \frac{2}{3}\text{Xe}+\frac{1}{3}{\text{XeO}}_3+4\text{HF}+\frac{1}{2}{\text{O}}_2$

From the stoichiometric coefficients, 1 mole of ${\text{XeF}}_4$ produces 4 moles of $\text{HF}$.

Therefore, the number of moles of HF produced by the complete hydrolysis of 1 mole of P is 4.