Balancing disproportionation reactions How to balance the following disproportionation reaction
 $$\ce {XeF2 + H2O → Xe + XeO3 + HF + O2} $$ 
 My attempt: I tried by making total number of reductions equal to total number of oxidation but I couldn't do because Xe is both oxidised and reduced also the solvent is water so what should I do to balance  protons and oxygen on both sides?.  Please help by solving it on the basis of same procedure that is making oxidation equal to reduction
It can't be balanced unambiguously, for this is not a single reaction, but a mechanical sum of two reactions, and these can be formally mixed in arbitrary proportion.

The first reaction is indeed a disproportionation. To balance it, use the common trick: consider these two xenons different elements for a while.
$$\ce{\underbrace{XeF2}_{for\;reduction} + \underbrace{XeF2}_{for\;oxidation} +H2O-&gt;Xe + XeO3 + HF}$$
The second is a typical redox:
$$\ce{XeF2 + H2O -&gt; Xe + O2 + HF}$$