In alkaline solution HXeO4- disproportionates: 2HXeO4- + 2OH- -> XeO6^4- + Xe + O2 + 2H2O. If 0.60 mol of HXeO4- reacts completely, the mass of xenon gas liberated is (Xe = 131)

JEE Advanced levelSingle correct optionDifficulty 3 of 3ph147

In alkaline solution $HXeO X_{4} X^{-}$ disproportionates: $2 HXeO X_{4} X^{-} + 2 OH X^{-} XeO X_{6} X^{4 -} + Xe + O X_{2} + 2 H X_{2} O$ . If $0.60 mol$ of $HXeO X_{4} X^{-}$ reacts completely, the mass of xenon gas liberated is (Xe = 131)

The balanced equation gives 1 mol of Xe per 2 mol of

HXeO X_{4} X^{-}

(Xe(VI) goes to Xe(VIII) and Xe(0)). Moles of Xe =

0.60/2 = 0.30

. Mass =

0.30 \times 131 = 39.3 g