Chemistry · Chemical equations and stoichiometry
Xenon reacts with fluorine at and 7 bar to form In this reaction the ratio of Xe
Xenon reacts with fluorine at \( 873 \mathrm{K} \) and 7 bar to form \( X e F_{4} . \) In this reaction the ratio of Xenon and fluorine required is :
- A. 1: 5
- B. 10:
- C. 1: 3
- D. 5:
Step-by-step solution
The balanced equation for the formation of XeF4 is Xe + 2F2 → XeF4, which gives a stoichiometric ratio of 1:2 (Xe:F2). However, in practice, to ensure complete conversion and suppress side products, an excess of fluorine is used. At 873 K and 7 bar, the typical molar ratio of xenon to fluorine is 1:5, as commonly reported for the synthesis of XeF4.
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