Identify the xenon compound formed with $\mathrm{O_2F_2}$

Dioxygen difluoride is a strong fluorinating agent. Xenon reacts with $\mathrm{O_2F_2}$ to form xenon tetrafluoride:

$\mathrm{Xe + 2O_2F_2 \rightarrow XeF_4 + 2O_2}$

So, the compound $\mathbf{P}$ is:

$\mathbf{P} = \mathrm{XeF_4}$

Write the complete hydrolysis reaction of $\mathrm{XeF_4}$

Complete hydrolysis of xenon tetrafluoride gives xenon trioxide, HF, and oxygen:

$6\mathrm{XeF_4} + 12\mathrm{H_2O} \rightarrow 2\mathrm{XeO_3} + 24\mathrm{HF} + 4\mathrm{Xe} + 3\mathrm{O_2}$

Calculate moles of HF produced per mole of $\mathrm{XeF_4}$

From the balanced equation:

$6\text{ mol } \mathrm{XeF_4} \rightarrow 24\text{ mol } \mathrm{HF}$

Therefore,

$1\text{ mol } \mathrm{XeF_4} \rightarrow \frac{24}{6} = 4\text{ mol } \mathrm{HF}$

Final answer

The number of moles of $\mathrm{HF}$ produced by complete hydrolysis of $1$ mol of $\mathbf{P}$ is:

$\boxed{4}$