1.

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

Answer»

`XeO_(3)`
`XeO_(2)F_(2)`
`XeOF_(4)`
`XE + XeO_(3)`

Solution :(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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