What is the best approach to balance this equation?
$$\ce{CO2 + H2O <--> C2H2 + O2}$$

I did it at first this way and I was not successful: \begin{align} \ce{4CO2 + H2O &<--> C2H2 + 3O2}\\ \ce{2CO2 + 6H2O &<--> 6C2H2 + 5O2}\\ \ce{12CO2 + 6H2O &<--> 6C2H2 + 5O2}\\ \end{align}

Should I start from right side or left? They say start from the most complicated molecule? If it is like that, then what is the most complicated one and why?

  • 1
    $\begingroup$ @airhuff Please do not add the reaction tag any more. We are in the process of blacklisting it. See meta for more information. $\endgroup$ Mar 29, 2017 at 7:15
  • $\begingroup$ Related: chemistry.stackexchange.com/questions/68924/… (Personally, I'm biased toward my own answer to this question, but I think all of them address how to balance a reaction in general without guesswork). $\endgroup$
    – Tyberius
    Mar 29, 2017 at 15:32

3 Answers 3


This may be a bit dumb method, but sometimes I do use it as it is pretty straightforward and doesn't require any guessing. I setup a few equations and try solving over integers. $$\ce{aCO2 + bH2O <--> cC2H2 + dO2}$$

Balancing $\ce{C}$ atoms both sides, $$a = 2c\\$$ Balancing $\ce{H}$ atoms both sides, $$2b=2c\\$$ Balancing $\ce{O}$ atoms both sides,


Solving gives this. Not hard to see that $a=4$ does the job.


The key for this type of equation is to balance the diatomic molecule last, and then just add a 1/2 of the diatomic molecule to finish the balance. Then, double all the coefficients and you're done.

  1. balance everything but the diatomic molecule: $$\ce{2CO2 + H2O <=> C2H2 + O2}$$

  2. Then add the right fraction of the diatomic molecule: $$\ce{2CO2 + H2O <=> C2H2 + 2.5O2}$$

Notice that 2.5 oxygen molecules is 5 oxygen atoms.

  1. Double all the coefficients to get whole numbers $$\ce{4CO2 + 2H2O <=> 2C2H2 + 5O2}$$

There is a standard equation for balancing such reactions: $$\ce{C_xH_y + [x +(y/4)]O2 -> x CO2 + y/2 H2O}$$

Hopefully, now you should be able to apply it to your reaction. It's just the reverse reaction that you're dealing with.

  • $\begingroup$ Please don't use shorthand or text speak. $\endgroup$ Mar 29, 2017 at 7:19
  • $\begingroup$ This answer would be more helpful if it demonstrated how the stoichiometric coefficients are found in the first place. $\endgroup$ Mar 29, 2017 at 7:52

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