XeF4 disproportionates in water: 6XeF4 + 12H2O -> 4Xe + 2XeO3 + 24HF + 3O2. If 0.60 mol of XeF4 reacts completely, and XeO3 and HF stay in solution, the number of moles of gas (Xe and O2) evolved is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ph212

$XeF X_{4}$ disproportionates in water: $6 XeF X_{4} + 12 H X_{2} O 4 Xe + 2 XeO X_{3} + 24 HF + 3 O X_{2}$ . If $0.60$ mol of $XeF X_{4}$ reacts completely, and $XeO X_{3}$ and $HF$ stay in solution, the number of moles of gas ( $Xe$ and $O X_{2}$ ) evolved is

For 6 mol

XeF X_{4}

the gases are 4 mol

Xe

plus 3 mol

O X_{2}

, which is 7 mol. For 0.60 mol, the scale factor is

0.60/6 = 0.10

, so gases

= 0.10 \times 7 = 0.70 mol

. The other products,

XeO X_{3}

and

HF

, remain dissolved.