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0.1 mole of XeF(6) is treated with 1.8 g...

0.1 mole of `XeF_(6)` is treated with 1.8 g of water.The product obtained is

A

`XeO_(3)`

B

`XeO_(2)F_(2)`

C

`XeOF_(4)`

D

`Xe + XeO_(3)`

Text Solution

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The correct Answer is:
C
(c) : `n_(H_(2)O) = (1.8)/(18) = 0.1` mole, `n_(XeF_(6)) = 0.1` mole
`XeF_(6) + H_(2)O to XeOF_(4) + 2HF`
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