Xe_{(g)} + 2 F_{2(g)} xrightarrow[7 text{ bar | vedclass

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(B) Xenon reacts with fluorine under different conditions to form $XeF_2$,$XeF_4$,and $XeF_6$.The reaction for the formation of $XeF_4$ is given by:$X

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$1 : 5$

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(B) Xenon reacts with fluorine under different conditions to form $XeF_2$,$XeF_4$,and $XeF_6$.The reaction for the formation of $XeF_4$ is given by:$Xe_{(g)} + 2 F_{2(g)} \xrightarrow{873 	ext{ K}, 7 	ext{ bar }} XeF_{4(s)}$In this reaction,$1 	ext{ mole}$ of $Xe$ reacts with $2 	ext{ moles}$ of $F_2$ to form $XeF_4$.However,in the industrial preparation of $XeF_4$,a mixture of $Xe$ and $F_2$ in a $1 : 5$ ratio is used to ensure the complete conversion of $Xe$ to $XeF_4$ and to prevent the formation of $XeF_2$ or $XeF_6$ as impurities.Therefore,the required ratio is $1 : 5$.

$Xe_{(g)} + 2 F_{2(g)} \xrightarrow[7 \text{ bar }]{873 \text{ K}} XeF_{4(s)}$
The ratio of $Xe : F_2$ required in the above reaction is

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$Xe_{(g)} + 2 F_{2(g)} \xrightarrow[7 \text{ bar }]{873 \text{ K}} XeF_{4(s)}$The ratio of $Xe : F_2$ required in the above reaction is