Xenon reacts with fluorine at 873 K and 7 bar to form XeF 4. In this reaction, the ratio of xenon and fluorine required is

Question

Xenon reacts with fluorine at 873 K and 7 bar to form XeF 4. In this reaction, the ratio of xenon and fluorine required is (A) 1: 5 (B) 10: 1 (C) 1: 3 (D) 5: 1

Tardigrade Admin · Accepted Answer

XeF 4 is obtained by heating a mixture of xenon and fluorine in the molar ratio of 1: 5 at 873 K and 7 bar pressure in an enclosed nickel vessel for a few hours. The reaction proceeds as Xe (g)+2 F 2(g) ->[873 K ][7 textbar] XeF 4 The extra fluorine taken, increases the production.

Q. Xenon reacts with fluorine at $873 K$ and $7$ bar to form $X e F_{4}$ . In this reaction, the ratio of xenon and fluorine required is

1 : 5

10 : 1

1 : 3

5 : 1

$X e F_{4}$ is obtained by heating a mixture of xenon and fluorine in the molar ratio of $1 : 5$ at $873 K$ and $7$ bar pressure in an enclosed nickel vessel for a few hours. The reaction proceeds as

$X e (g) + 2 F_{2} (g) 873 K 7 bar X e F_{4}$

The extra fluorine taken, increases the production.

Q. Xenon reacts with fluorine at 873K and 7 bar to form XeF4​. In this reaction, the ratio of xenon and fluorine required is