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I can see that denitrification consists of processes involving mainly two steps:

Step 1 (nitrate to nitrite): $$ \ce{3NO3- + CH3OH -> 3NO2- + CO2 + 2H2O} $$

Step 2 (nitrite to molecular nitrogen): $ \ce{2NO2- + CH3OH -> N2 + CO2 + H2O + 2OH-} $$

Total reaction is supposed to be:

$$ \ce{6NO3- + 5CH3OH -> 3N2 + 5CO2 + 7H2O + 6 OH-} $$

Now I'm a bit rusty.

As far as I can remember, the right hand side on the first step of the reaction will be the starting point for the left hand side of step 2 in the overall reaction process.

In other words, what is present on the right hand side of step 1 and on the left hand side of step 2 can be subtracted from each other (which is logical). Also, the start of the total reaction would be represented by the left hand side of step 1 and the end of the total reaction would be represented by the right hand side of step 2.

The coefficients 5 and 7 seem very strange, but they balance everything so that makes sense. However the balancing of the Os seem to be the challenging part.


The attempt yields:

$$ \ce{3 NO3- + CH3OH -> N2 + CO2 + H2O + 2OH-} $$

Balancing N:

$$ \ce{6NO3- + CH3OH -> 3N2 + CO2 + H2O + 2OH-} $$

C and H atoms are already balanced.

However balancing O is more complicated because changing any of the Os affects the balance of the others (like C, H). Are there any recommendations for easing out that process or methods that can be applied in general?

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After balancing carbon and nitrogen, I'd recommend you go for balancing the charge in the equation:

$$ \ce{6NO3- + CH3OH -> 3N2 + CO2 + H2O + 6OH-} $$

Now, we've to balance oxygen and hydrogen without disturbing the already balanced species. It would be interesting to note that here the stoichiometric coefficients of methanol and carbon dioxide should be the same or the number of carbon atoms wouldn't be equal on both sides of the equation.

So, let's assume the stoichiometric coefficient for both these two species to be $x$. Also, let's assume the stoichiometric coefficient of water on the right hand side to be $y$. The equation now looks like this:

$$ \ce{6NO3- + xCH3OH -> 3N2 + xCO2 + yH2O + 6OH-} $$

We now balance the number of oxygen atoms:

$$ 18 + x = 2x + y + 6 $$

And the number of hydrogen atoms:

$$ 4x = 2y + 6 $$

Solving these two equations, you get $x=5$ and $y=7$. And therefore your overall equation finally is:

$$ \ce{6NO3- + 5CH3OH -> 3N2 + 5CO2 + 7H2O + 6OH-} $$

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  • $\begingroup$ That definitely eases the process to a high degree, thanks! However regarding the assumptions for the stoichiometric coefficients that you use where you define the stoichiometric coefficient of water to be equal to y: is it normal to treat water in this way when balancing chemical reactions? Or is it rather so that it is not a rule of thumb, but something one finds by trial and error and/or experience? So perhaps this doesn't only apply to water. $\endgroup$ – jibo May 13 '17 at 10:41
  • $\begingroup$ @jibo It just struck me that this was the quickest way I could get to the answer, I'm aware that this is really a vague comment, but now you're interested I'll write a canonical post about balancing equations. It would then make your concepts crystal clear. Please allow me some time. $\endgroup$ – Berry Holmes May 13 '17 at 17:22

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