Xe_{(g)} and F_{2(g)} react in the ratio 1:20 | vedclass

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(B) The reaction of $Xe$ with $F_2$ in a $1:20$ molar ratio at $573 \ K$ and $60-70 \ bar$ pressure yields $XeF_6$ $(A)$.$Xe_{(g)} + 3F_{2(g)} \xright

Vedclass · Accepted Answer

$XeF_6, XeO_3$

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(B) The reaction of $Xe$ with $F_2$ in a $1:20$ molar ratio at $573 \ K$ and $60-70 \ bar$ pressure yields $XeF_6$ $(A)$.$Xe_{(g)} + 3F_{2(g)} \xrightarrow{573 \ K, 60-70 \ bar} XeF_{6(s)}$Complete hydrolysis of $XeF_6$ produces $XeO_3$ $(B)$ and $HF$.$XeF_{6(s)} + 3H_2O_{(l)} \rightarrow XeO_{3(s)} + 6HF_{(aq)}$Therefore,$A$ is $XeF_6$ and $B$ is $XeO_3$.

$Xe_{(g)}$ and $F_{2(g)}$ react in the ratio $1:20$ at $573 \ K$ and $60-70 \ bar$ to form $A$. When $A$ is completely hydrolyzed,$B$ and $HF$ are formed. $A$ and $B$ are respectively:

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