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The reaction XeF4 + H2O gives Xe + O2 + HF + XeO3 can be balanced in more than one way and I can not understand why. Please help.

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    $\begingroup$ Check here and the other answers there as well. $\endgroup$ – Ed V Jul 10 '20 at 17:48
  • $\begingroup$ Actually, if the cited reaction produces transient radical species, I would postulate that it is possible that the reaction can NOT be balanced, with respect to measured observed products, in a single reaction equation. Examples, the action of HNO3 on Copper metal. Unclear if this is the intent of the question, a clue would be at what level is the science being studied. $\endgroup$ – AJKOER Jul 10 '20 at 21:37
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    $\begingroup$ Please consider giving an upvote to the most helpful of the posted answers either at the link I gave or at one of the linked duplicates. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this! $\endgroup$ – Ed V Jul 10 '20 at 23:25
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There is a quicker way to solve this equation, without any algebra. It is to imagine that, first $\ce{XeF_4}$ is hydrolyzed by $\ce{H_2O}$ forming $\ce{HF}$ and a hypothetical xenon oxide without any change of oxidation number. And secondly this hypothetical oxide $\ce{XeO_2}$gets disproportionized into $\ce{Xe}$ and $\ce{O_2}$, or into $\ce{Xe + XeO_3}$ $$\ce{XeF_4 + 2 H_2O -> XeO_2 + 4 HF} .......\ (1)$$ $$\ce{XeO_2 -> Xe + O_2} ....................(2) $$ or $$\ce{3 XeO_2 -> Xe + 2 XeO_3} \ ..............(3) $$

The simplest solution is adding ($2$) and ($3$), and then adding $4$ times ($1$). $\ce{XeO_2}$ disappears, and it gives : $$\ce{4 XeF_4 + 8 H_2O -> 16 HF + 2 Xe + 2 XeO_3 + O_2}$$

But there is an infinite number of other solutions. Multiplying ($2$) by x, ($3$) by y, and ($1$) by x + 3y, then adding the obtained equations, you get the general solution of the equation :

$$\ce{(x + 3y) XeF_4 + 2(x + 3y) H_2O -> 4(x + 3y) HF + (x + y) Xe + 2y XeO_3 + x O_2}$$

For example, choosing x = $1$ and y = $10$, one gets $$\ce{31 XeF_4 + 62 H_2O -> 124 HF + 11 Xe + 20 XeO_3 + O_2}$$

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  • $\begingroup$ Per books.google.com/… also possible F2, no O2. Adding H2O forms HF, and O2 or O3, see chemistryscl.com/reactions/fluorine+water-reaction/…. $\endgroup$ – AJKOER Jul 10 '20 at 22:36
  • $\begingroup$ Maurice, per my last two references, the reaction system is complex with varying products depending on the amount of water present. Also possible XeOF2. Your cited reactions thus appear to be largely speculations unless you have a source, $\endgroup$ – AJKOER Jul 10 '20 at 22:42
  • $\begingroup$ @ Ajkoer. You are right. My method is a speculation. I have told it : it is hypothetical. It has the advantage of being able to quickly solve the problem. And of course you could propose XeOF2 as intermediate. $\endgroup$ – Maurice Jul 11 '20 at 9:34

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