When $262 g$ of xenon (atomic mass $= 131$ ) reacted completely with $152 g$ of fluorine (atomic mass $= 19)$ , a mixture of $X e F_{2}$ and $X e F_{6}$ was produced. The molar ratio $X e F_{2} : X e F_{6}$ is :

1 : 2

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1 : 4

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1 : 1

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1 : 3

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Solution

The correct option is C $1 : 1$
$X e \frac{262}{131} + F_{2} \frac{152}{38} \to X e F_{2} a m o l e + X e F_{6} b m o l e$

Let $a$ mole $X e F_{2}$ form and $b$ mole $X e F_{6}$ from.

Apply $P O A C$ ,
$a \times 1 + b \times 1 = 2$ ------- $(1)$
$2 a + 6 b = 8$ --------- $(2)$

After solving eq. $(1)$ and $(2)$
$a = 1 m o l e$ and $b = 1 m o l e$ .

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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 :

When Xenon reacts with fluorine in a ratio of 1:5 at a temperature of 873 K it forms:

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If true enter 1, else enter 0.

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When 262 g of xenon (atomic mass =131) reacted completely with 152 g of fluorine (atomic mass =19), a mixture of XeF2 and XeF6 was produced. The molar ratio XeF2:XeF6 is :

The correct option is C 1:1Xe262131+F215238→XeF2a mole+XeF6b moleLet a mole XeF2 form and b mole XeF6 from.Apply POAC,a×1+b×1=2 ------- (1)2a+6b=8 --------- (2)After solving eq. (1) and (2)a=1 mole and b=1 mole.

When $262 g$ of xenon (atomic mass $= 131$ ) reacted completely with $152 g$ of fluorine (atomic mass $= 19)$ , a mixture of $X e F_{2}$ and $X e F_{6}$ was produced. The molar ratio $X e F_{2} : X e F_{6}$ is :