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Consider the following unbalanced reaction:

$$\ce{XeF4 + H2O -> Xe + XeO2 + HF + O2}$$

There are six variables to be solved, but only four equations linking them. I solved the set of equations using Gauss elimination and arbitrarily set the stoichiometric coefficient of $\ce{XeF4}$ as $3$:

$$\ce{3XeF4 + 6H2O -> $(4k)$Xe + $(3-4k)$XeO2 + 12HF + $(4k)$O2}$$

where $k$ is the free variable, the parameter, with $0<k<\frac 34$ (since coefficients must be positive).

Clearly, for different values of $k$, it gives radically different yet balanced equations. The very yields are different for fixed amount of $\ce{XeF4}$ and $\ce{H2O}$. This issue has been noted in a previous question.

My textbook, however, gives the balanced equation:

$$\ce{6XeF4 + 12H2O -> 4Xe + 2XeO2 + 24HF + 4O2}$$

clearly setting $k=\frac 12$. The linked question above does not offer any explanation for why a particular value of $k$ should be favoured. So, my exact question is: Why set $k=\frac 12$? Why was $\frac 12$ preferred over any other value for $k$:

  • Is it determined by experimentally measuring the stoichiometry?
  • Is there a theoretical explanation for it?
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    $\begingroup$ related chemistry.stackexchange.com/questions/37684/… $\endgroup$
    – Mithoron
    Apr 23, 2017 at 18:29
  • $\begingroup$ chemical equations are not unlike math equations, multiplying each side by same number will give a valid equation. Similarly, addition of two (or more) equations side-wise with subsequent simplification gives a valid equation. However, if the equation depicts an actual chemical process is another matter, a well-balanced equation may be irrelevant for this reason. $\endgroup$
    – permeakra
    Apr 23, 2017 at 23:41
  • $\begingroup$ @Mithoron In the answer to that question, they just say that the balanced equations are different. I want to know which is correct and why? Which factors determine that? And it is not a duplicate because he mentions just 2 possibilities and I found all the solutions. $\endgroup$ Apr 24, 2017 at 3:12
  • $\begingroup$ I would change the title of your question to better reflect the change you made. Something along the lines of "How to determine true reaction stoichiometry among multiple valid options". $\endgroup$
    – Tyberius
    Apr 24, 2017 at 3:41
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    $\begingroup$ There is a meta discussion on the closure of this question: Possible overreach on marked duplicate $\endgroup$ Apr 24, 2017 at 16:00

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