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JEE Advanced levelSingle correct optionDifficulty 3 of 3ph145

$0.10 mol$ of $XeF X_{6}$ reacts completely with exactly $0.25 mol$ of water, with all the water consumed. The products are only $XeOF X_{4}$ and $XeO X_{3}$ (with HF). The amount of $XeO X_{3}$ formed is

A0.025 mol
B0.050 mol
C0.075 mol
D0.100 mol

Answer: C

Let

a

mol give

XeOF X_{4}

(

XeF X_{6} + H X_{2} O XeOF X_{4} + 2 HF

) and

b

mol give

XeO X_{3}

(

XeF X_{6} + 3 H X_{2} O XeO X_{3} + 6 HF

). Then

a + b = 0.10

and

a + 3 b = 0.25

. Subtracting gives

2 b = 0.15

, so

b = 0.075 mol

(and

a = 0.025

).

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