XeF6 is completely hydrolysed by water: XeF6 + 3H2O -> XeO3 + 6HF. Take Xe = 131, F = 19, O = 16, so the molar masses are XeF6 = 245 and XeO3 = 179 g mol^-1. Calculate the mass of XeO3 formed from 3.71 g of XeF6, in grams to two decimal places.

JEE Main levelNumerical answerDifficulty 2 of 3ph211

$XeF X_{6}$ is completely hydrolysed by water: $XeF X_{6} + 3 H X_{2} O XeO X_{3} + 6 HF$ . Take Xe = 131, F = 19, O = 16, so the molar masses are $XeF X_{6}$ = 245 and $XeO X_{3}$ = 179 $g mo l^{- 1}$ . Calculate the mass of $XeO X_{3}$ formed from $3.71 g$ of $XeF X_{6}$ , in grams to two decimal places.

Numerical answer type, answer in g, tolerance 0.01.

n (XeF X_{6}) = 3.71/245 = 0.01514 mol

. The equation gives 1 mol

XeO X_{3}

per mol

XeF X_{6}

. Mass of

XeO X_{3} = 0.01514 \times 179 = 2.71 g