$\mathrm{Xe}(\mathrm{g})+2 \mathrm{~F}_2(\mathrm{~g}) \xrightarrow[7 \text { bar }]{873}…
$\mathrm{Xe}(\mathrm{g})+2 \mathrm{~F}_2(\mathrm{~g}) \xrightarrow[7 \text { bar }]{873} \mathrm{XeF}_4(\mathrm{~s})$
The ratio of $\mathrm{Xe}: \mathrm{F}_2$ required in the above reaction is
$1: 2$
$1: 5$
$1: 20$
$1: 12$
Solution
Xenon forms three binary fluorides $\mathrm{XeF}_2, \mathrm{XeF}_4$ and $\mathrm{XeF}_6$ by direct reaction of elements under appropriate conditions.
$\mathrm{Xe}(\mathrm{~g})+2 \mathrm{~F}_2(\mathrm{~g}) \xrightarrow{873 \mathrm{~K}, 7 \text { bar }} \mathrm{XeF}_4(\mathrm{~s})$
(1:5 ratio)