If $\mathrm{K}_1$ and $\mathrm{K}_2$ are the respective equilibrium constants for the two reactions,
$$
\mathrm{XeF}_6(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{~g}) \rightleftharpoons \mathrm{XeOF}_4(\mathrm{~g})+2 \mathrm{HF}(\mathrm{~g})
$$
$\mathrm{XeO}_4(\mathrm{~g})+\mathrm{XeF}_6(\mathrm{~g}) \rightleftharpoons \mathrm{XeOF}_4(\mathrm{~g})+\mathrm{XeO}_3 \mathrm{~F}_2(\mathrm{~g})$
Then equilibrium constant of the reaction $\mathrm{XeO}_4(\mathrm{~g})+$ $2 \mathrm{HF}(\mathrm{g}) \rightleftharpoons \mathrm{XeO}_3 \mathrm{~F}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ will be
(a) $\mathrm{K}_1 /\left(\mathrm{K}_2\right)^2$
(b) $\mathrm{K}_1 \cdot \mathrm{~K}_2$
(c) $\mathrm{K}_1 / \mathrm{K}_2$
(d) $\mathrm{K}_2 / \mathrm{K}_1$