1. Tardigrade
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  3. Chemistry
  4. Xe react directly with F2 in different conditions by volume. Ratio of Xe and F2 is given to make different Xenon fluoride products. XeF2⇒ Xe:F2=x:a=1:0.5 XeF4⇒ Xe:F2=x:b XeF6⇒ Xe:F2=x:c The value of a+b+c will be Round off your answer to the nearest integer after multiplying with 10.

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Q. $Xe$ react directly with $F_{2}$ in different conditions by volume. Ratio of $Xe$ and $F_{2}$ is given to make different Xenon fluoride products.
$XeF_{2}\Rightarrow Xe:F_{2}=x:a=1:0.5$
$XeF_{4}\Rightarrow Xe:F_{2}=x:b$
$XeF_{6}\Rightarrow Xe:F_{2}=x:c$
The value of $a+b+c$ will be
Round off your answer to the nearest integer after multiplying with $10$.

NTA AbhyasNTA Abhyas 2022

Solution:

For $XeF _{2} \Rightarrow Xe _{2} F _{2}=1: 0.50$
For $XeF _{4} \Rightarrow Xe : F _{2}=1: 5$
For $XeF _{6} \Rightarrow Xe : F _{2}=1: 20$
Sum $=20+5+0.50=25.50$