When 262 g of xenon (atomic mass = 131) reac | vedclass

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(C) The chemical reaction is: $2Xe + 4F_2 \longrightarrow XeF_2 + XeF_6$Initial moles of $Xe = \frac{262}{131} = 2 \ mol$.Initial moles of $F_2 = \fra

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$1 : 1$

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(C) The chemical reaction is: $2Xe + 4F_2 \longrightarrow XeF_2 + XeF_6$Initial moles of $Xe = \frac{262}{131} = 2 \ mol$.Initial moles of $F_2 = \frac{152}{38} = 4 \ mol$.Let $x$ be the moles of $XeF_2$ and $y$ be the moles of $XeF_6$ formed.According to the law of conservation of mass (atoms):For $Xe$: $x + y = 2$For $F$: $2x + 6y = 2 	imes 4 = 8$Solving the equations: $x + y = 2$ and $x + 3y = 4$.Subtracting the first from the second: $2y = 2$,so $y = 1$.Substituting $y = 1$ into $x + y = 2$,we get $x = 1$.Thus,the molar ratio $XeF_2 : XeF_6$ is $1 : 1$.

When $262 \ g$ of xenon (atomic mass $= 131$) reacted completely with $152 \ g$ of fluorine (atomic mass $= 19$),a mixture of $XeF_2$ and $XeF_6$ was produced. The molar ratio $XeF_2 : XeF_6$ is

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