# How come the same reaction can be balanced in two distinct ways?

I faced this reaction which can be balanced in $2$ distinct ways which are not multiples of each other.

\begin{align}\ce{6XeF4 + 12H2O &-> 2XeO3 + 24HF + 4Xe + 3O2}\tag{1}\\[0.4em] \ce{4XeF4 + 8H2O &-> 2XeO3 + 16HF + 2Xe + O2}\tag{2}\end{align}

The first reaction is as it is given in the books and the second reaction was balanced by me.

I sense that only a mechanism can resolve this. I found a similar question on googling, but the answer is unsatisfactory. (It's Yahoo Answers. What can you expect?)

I also found two independent reactions here which on adding produce equation $(2)$. I am confused as to what exactly happens.

• Interesting. Try this out for fun: treat those chemical equations exactly like you would mathematical equations. Subtract the second equation from the first (multiply the equations by constants if necessary) and try to cancel out as many substances as possible (i.e. give them stoichiometric coefficients of 0). If you find negative coefficients, don't be alarmed; you can turn a negative coefficient into a positive one by swapping it from the left-hand side to the right or vice-versa (just like in a mathematical equation, where the reaction arrow represents an equal sign). What do you find? May 22 '14 at 13:23
• @NicolauSakerNeto I tried. I got some reactions. But that was a long time ago. Do you want me to post that in the question? May 22 '14 at 13:33
• You don't necessarily have to add it to the question. All I mean to say is that if you play with the equations a bit, the answer may become clear. I'd love to give you a deep answer myself, but I lack the time to do so presently. If no one else answers though, I'll be happy to give a short but direct response in about 12 hours. If you do figure it out, though, feel free to answer your own question! May 22 '14 at 13:37
• @NicolauSakerNeto Is my reasoning in my answer correct? May 22 '14 at 13:48

As per Nicolau's suggestion, :

$$\{(1-2)\times\color\red{4}\}-2$$

gives

$$2\ce{XeO3}\to2\ce{Xe}+\ce{3O2}$$

Thus, I conjecture that the second and third reaction differ by the fact of decomposition of $\ce{XeO3}$

$2$ can be converted into $1$ by multiplying by $\color \red{\frac 3 2}$ and decomposing one of the $\ce{XeO3}$ molecules.

Wikipedia also says that $\ce{XeO3}$ is unstable.

• 3 times equation 2 can be converted to 2 times equation 1, by decomposing 2 XeO3 molecules in 3 times equation 2. May 22 '14 at 19:26
• Precisely the reaction you had to find! Good job, though I think you missed multiplying reaction $2$ by 2 outside the brackets, and it is reaction $2$ that can be tuned into $1$ by decomposing one $\ce{XeO3}$ for every six $\ce{XeF4}$ . Stoichiometry alone cannot pick the "right" equation of the two. The reaction mechanism and conditions would have to determine which is a better description. With no additional knowledge, perhaps equation $2$ is slightly better, as one might be asked to justify why only a third of the $\ce{XeO3}$ molecules decompose in equation $1$. May 22 '14 at 22:15
• Oh yeah... could you please provide a better description? May 23 '14 at 3:13
• I'm not sure how I could improve on your answer, really. Do you have a specific question? May 23 '14 at 23:18
• @NicolauSakerNeto You said : " The reaction mechanism and conditions would have to determine which is a better description" May 24 '14 at 3:23