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I found this equation while testing my equation balancer: $$ \ce{a C7H6O3 + b C4H6O3 -> c C9H8O4 + d C2H4O2} $$

Obviously, $a = b = c = d = 1$ is a solution. However, if we were to bring in some maths:

$ \begin{cases} 7 a + 4b &= 9c + 2d\\ 6a + 6b &= 8c + 4d\\ 3a + 3b &= 4c + 2d\\ \end{cases}\\ \begin{cases} a &= \frac{11}{9}\alpha - \frac{2}{9}\beta\\ b &= \frac{1}{9}\alpha + \frac{8}{9}\beta\\ c &= \alpha\\ d &= \beta\\ \end{cases} $

This leaves us with many ways of balancing the equation:

  • when $\alpha = 1, \beta = 1$, $a = b = c = d = 1$; $\ce{C7H6O3 + C4H6O3 -> C9H8O4 + C2H4O2}$
  • when $\alpha = 1, \beta = 2$, $\begin{cases} a &= 7\\ b &= 17\\ c &= 9\\ d &= 18\\ \end{cases}$; $\ce{7 C7H6O3 + 17 C4H6O3 -> 9 C9H8O4 + 18 C2H4O2}$
  • when $\alpha = 2, \beta = 1$...

From what I know, this happens when the equation consists of 2 or more independent equations (as is the case here: More than one way of balancing a chemical equation). However, here the two equations are:

$$ \ce{11 C7H6O3 + C4H6O3 -> 9 C9H8O4} \\ \ce{-2 C7H6O3 + 8 C4H6O3 -> 9 C2H4O2} $$

What is concerning here is that the second equation has a negative stoichiometric coefficient. My question is, is a negative coefficient acceptable in an equation, and in this case which is the correct way to balance the equation?

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  • $\begingroup$ Your equation was already balanced, so why would you try to get other solution? $\endgroup$ – Mithoron May 1 '18 at 19:25
  • $\begingroup$ @Mithoron I guess I am being more a mathematician than a chemist here, but clearly there are other solutions, which makes me wonder if they too are valid solutions. $\endgroup$ – Jingjie YANG May 1 '18 at 19:32
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    $\begingroup$ There is nothing special about a negative coefficient. If you prefer not to have it, transfer that term to the other side. $\endgroup$ – Ivan Neretin May 1 '18 at 19:51
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    $\begingroup$ You may answer your question yourself. $\endgroup$ – Ivan Neretin May 1 '18 at 20:17
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    $\begingroup$ @Jingjie YANG you are correct, negative coefficients just flip to be positive on the other side. $\endgroup$ – Tyberius May 2 '18 at 1:01
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As @Ivan Neretin and @Tyberius pointed out, the negative coefficient of a reactant can be flipped to be positive by moving the reactant to the products' side.

So, in my case, the equations become $$\begin{align} \ce{11C7H6O3 + C4H6O3 &-> 9C9H8O4}\\ \ce{8C4H6O3 &-> 2C7H6O3 + 9C2H4O2} \end{align}$$

Now, since the original equation $\ce{C7H6O3 + C4H6O3 -> C9H8O4 + C2H4O2}$ consists of 2 independent equations which can be separately balanced and combined in an arbitrary proportion. See More than one way of balancing a chemical equation for a more detailed explanation.

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