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Xenon reacts with fluorine at $873 \mathrm{~K}$ and 7 bar to form $\mathrm{XeF}_4$. In this reaction, the ratio of xenon and fluorine required is
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$1: 5$
$\mathrm{XeF}_4$ is obtained by heating a mixture of xenon and fluorine in the molar ratio of $1: 5$ at $873 \mathrm{~K}$ and 7 bar pressure in an enclosed nickel vessel for a few hours. The reaction proceeds as
$$
\mathrm{Xe}(g)+2 \mathrm{~F}_2(g) \stackrel{873 \mathrm{~K}}{7 \text { bar }} \mathrm{XeF}_4
$$
The extra fluorine taken, increases the production.
$$
\mathrm{Xe}(g)+2 \mathrm{~F}_2(g) \stackrel{873 \mathrm{~K}}{7 \text { bar }} \mathrm{XeF}_4
$$
The extra fluorine taken, increases the production.
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