Xe + 2F_2 xrightarrow{573 K, 60.70 bar} AW | vedclass

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(D) The reaction of Xenon $(Xe)$ with Fluorine $(F_2)$ depends on the molar ratio and reaction conditions.When $Xe$ reacts with $F_2$ in a $1:2$ molar

Vedclass · Accepted Answer

$XeF_4$

Vedclass · Answer

(D) The reaction of Xenon $(Xe)$ with Fluorine $(F_2)$ depends on the molar ratio and reaction conditions.When $Xe$ reacts with $F_2$ in a $1:2$ molar ratio at $573 \ K$ and $60-70 \ bar$ pressure,it produces Xenon tetrafluoride $(XeF_4)$.The balanced chemical equation is: $Xe(g) + 2F_2(g) \xrightarrow{573 \ K, \ 60-70 \ bar} XeF_4(s)$.Therefore,$A$ is $XeF_4$.

$Xe + 2F_2 \xrightarrow{573 \ K, \ 60.70 \ bar} A$
What is $A$?

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$Xe + 2F_2 \xrightarrow{573 \ K, \ 60.70 \ bar} A$What is $A$?