A 3.105 g sample of XeF4 is hydrolysed completely in water. The only xenon products are Xe and XeO3 (Xe(IV) disproportionates), and the only oxidation product of water is O2. Using a balanced equation, find the volume of O2 at STP, in mL (nearest integer). Atomic masses: Xe = 131, F = 19; molar volume at STP = 22.4 L/mol.

JEE Advanced levelNumerical answerDifficulty 3 of 3ph151

A $3.105 g$ sample of $XeF X_{4}$ is hydrolysed completely in water. The only xenon products are Xe and $XeO X_{3}$ (Xe(IV) disproportionates), and the only oxidation product of water is $O X_{2}$ . Using a balanced equation, find the volume of $O X_{2}$ at STP, in mL (nearest integer). Atomic masses: Xe = 131, F = 19; molar volume at STP = 22.4 L/mol.

Numerical answer type, answer in mL, tolerance 1.

Balanced equation:

6 XeF X_{4} + 12 H X_{2} O 4 Xe + 2 XeO X_{3} + 24 HF + 3 O X_{2}

. Molar mass of

XeF X_{4}

131 + 76 = 207 g mo l^{- 1}

, so

n = 3.105/207 = 0.0150 mol

O X_{2}

formed =

0.0150 \times 3/6 = 0.00750 mol

. Volume =

0.00750 \times 22.4 = 0.168 L = 168 mL