Q. 0.1 mole of XeF6 is treated with 1.8 g of water. The product obtained is

Solution:

$XeF_{6}+H_{2}O \to XeOF_{4}+2HF$
$1 : 1\quad1$
0.1 mole of $XeF_{6} \, reacts \,with\, 0.1 mol \,of H_{2}O \left(1.8 g\right) \,to \,give \, XeOF_{4}$

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