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This question already has an answer here:

I am trying to balance some chemical reactions by using the Gauss' elimination method and I noticed that this method does not work if the creation of a matrix with exactly one degree of freedom is not possible.

In this example

$$\ce{H2O2 + NO3- + H+ → O2 + NO + H2O}$$

I succeeded in creating just 4 equations

$$\ce{\alpha H2O2 + \beta NO3- + \gamma H+ → \delta O_2 + \epsilon NO + \zeta H2O}$$

$$ \left\{ \begin{aligned} 2\alpha+\gamma&=2\zeta &\qquad &(\ce{H})\\ 2\alpha+3\beta&=2\delta+\epsilon+\zeta &\qquad &(\ce{O})\\ \beta&=\epsilon &\qquad &(\ce{N})\\ -\beta+\gamma&=0 &\qquad &(\text{electrons}) \end{aligned} \right. $$

obtaining

$$(\alpha, \beta, \gamma, \delta, \epsilon, \zeta)=\left(\zeta-\frac{1}{2}\epsilon, \epsilon,\epsilon,\frac{\zeta+\epsilon}{2}, \epsilon, \zeta\right)$$

I know that the solution should be $(3,2,2,3,2,4)$, but, in having two degrees of freedom, I am not able to choose the two free variables in order to obtain what I expect, i.e. $\epsilon=2$ and $\zeta=4$.

Is there any (just one more) equation that I missed? Is there anything else that I have to add further? Or do I have to consider that Gauss' elimination fails with problems with a number of free variables $\neq 1$?

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marked as duplicate by Karsten Theis, Tyberius, Mithoron, Poutnik, Jon Custer Jun 26 at 17:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Your "should-be" solution,

$$\ce{3H2O2 + 2NO3- + 2H+ → 3O2 + 2NO + 4H2O}$$

is a linear combination of

$$\ce{H2O2 + 2NO3- + 2H+ → 2O2 + 2NO + 2H2O}$$

and

$$\ce{2H2O2 → O2 + 2H2O}$$

Any other linear combination is valid as well.

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    $\begingroup$ I did first write this answer and then voted to close the question, which might be viewed as a mixed message. I'm happy to delete this answer if there are more down votes (1 up 1 down as I write this). $\endgroup$ – Karsten Theis Jun 26 at 13:15
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    $\begingroup$ There's no prob. with answer on duplicate. $\endgroup$ – Mithoron Jun 26 at 14:51

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