Xenon tetrafluoride undergoes complete…

Xenon tetrafluoride undergoes complete hydrolysis according to the (unbalanced) reaction XeF4+H2O→ Xe+XeO3+HF+O2, in which Xe and XeO3 are produced…

Single correct option

Xenon tetrafluoride undergoes complete hydrolysis according to the (unbalanced) reaction

\mathrm{XeF_4+H_2O\to Xe+XeO_3+HF+O_2}

, in which Xe and

\mathrm{XeO_3}

are produced in the mole ratio

4:2

. When

0.0100

mol of

\mathrm{XeF_4}

is hydrolysed completely, the volume of gas (Xe and

\mathrm{O_2}

together, measured at

273\ \mathrm{K}

and

1\ \mathrm{atm}

, molar volume

22.4\ \mathrm{L}

) liberated, assuming

\mathrm{XeO_3}

stays in solution, is

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Answer: (A) $0.26\ \mathrm{L}$

Balance with the given ratio

\mathrm{Xe:XeO_3}=4:2

. Xe atoms:

6\ \mathrm{XeF_4}

give

4\ \mathrm{Xe}+2\ \mathrm{XeO_3}

. Fluorine:

24

F go to

24\ \mathrm{HF}

, which needs

24

H from

12\ \mathrm{H_2O}

. Oxygen:

12

O on the left;

2\ \mathrm{XeO_3}

hold

6

O, so

6

O form

3\ \mathrm{O_2}

. The balanced equation is

\mathrm{6XeF_4+12H_2O\to4Xe+2XeO_3+24HF+3O_2}

(atoms check: Xe 6, F 24, H 24, O 12). Hence

6

mol

\mathrm{XeF_4}

liberates

4+3=7

mol of gas. For

0.0100

mol: gas

=\frac{7}{6}\times0.0100=0.01167

mol, volume

=0.01167\times22.4=0.261\ \mathrm{L}

. Counting only Xe would give

0.15\ \mathrm{L}

, and counting

\mathrm{XeO_3}

as a gas as well would give

\tfrac96\times0.0100\times22.4=0.34\ \mathrm{L}

Original question written in the JEE Advanced 2017 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake