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 _______.

Dioxygen difluoride, $$O_2F_2$$, is an extremely powerful oxidising as well as fluorinating agent. When it is allowed to react with xenon, it oxidises xenon to the +6 oxidation state while simultaneously supplying four fluorine atoms. The compound obtained is the oxy-fluoride $$XeOF_4$$ (called xenon oxytetrafluoride).

A convenient balanced equation for its formation is

$$2\,Xe + 4\,O_2F_2 \;\longrightarrow\; 2\,XeOF_4 + 3\,O_2$$

Let the product $$P$$ be $$XeOF_4$$. To find the number of moles of $$HF$$ produced on complete hydrolysis of one mole of $$P$$, write its hydrolysis reaction:

$$XeOF_4 + 2\,H_2O \;\longrightarrow\; XeO_3 + 4\,HF$$

Checking the atom balance: Xe (1), O (1 + 2 = 3 on both sides), F (4), and H (4) are all balanced, so the equation is correct.

The equation shows that 1 mole of $$XeOF_4$$ yields 4 moles of $$HF$$ on complete hydrolysis.

Hence, the required number of moles of $$HF$$ produced per mole of $$P$$ is 4.