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 :

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

Xe + F $_{2}$ $\to$ XeF $_{2}$ + XeF $_{6}$

$\frac{262}{131}$ $\frac{152}{38}$ a mole b mole
= 2 mole 4 mole
Let a mole XeF $_{2}$ form and b mole XeF $_{6}$ form
Apply POAC
$a \times 1 + b \times 1 = 2 \dots \dots$ (1)
2a + 6b = 8 $\dots \dots$ (2)
After solving eq. (1) & (2)
a = 1 mole & b = 1 mole

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Similar questions

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

How are xenon fluorides $X e F_{2}, X e F_{4}$ and $X e F_{6}$ obtained?

X e F_{6} > X e F_{4} > X e F_{2}

X

X e F_{2}, X e F_{4}

X e F_{6}

X e F_{2}

X e F_{6}

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