QJEEAdv22Chbe2a: JEE Advanced 2022 Chemistry - Class 12 JEEAdv22Chbe2a Solution: The reaction of and gives a compound . The number of moles of…

$$\mathrm{Xe}+2{\mathrm{O}}_2{F}_2\to \underset{(\mathrm{P})}{{\mathrm{XeF}}_4}+2{\mathrm{O}}_2$$

$$\underset{(\mathrm{P})}{{6\mathrm{XeF}}_4}+12{\mathrm{H}}_2\mathrm{O}⟶4\mathrm{Xe}+2{\mathrm{XeO}}_3+24\mathrm{HF}+3{\mathrm{O}}_2$$

So, from the above reaction, it is clear that 6 moles of $\mathrm{XeF}_4$ produces 24 moles of $\mathrm{HF}$ .

So, 1 mole of $\mathrm{XeF}_4$ will produce $\frac{24}{6}$ moles of HF, i.e., 4 moles of $\mathrm{HF}$ .