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Why does only one reaction take place, although so many would be possible? I'm more of a physics and math guy, so simply "accepting" it without a proper explanation is kind of hard for me. For example: $$\ce{Na2CO3 + HCl -> NaCl + H2O + CO2}.$$

Why can't $\ce{H+}$ combine with $\ce{CO3^2-}$ to form a compound?

I'm going to highschool, and I am great at physics and mathematics, and I dislike chemistry since it's mostly memorizing. Please describe it in terms I can understand. Also, this is something that I genuinely want to understand rather than memorize, it is not related to school.

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    $\begingroup$ $\ce{H+}$ not only can, but actually does combine with $\ce{CO3^2-}$ to form a compound. The problem is what happens next. Also, welcome to Chem.SE. $\endgroup$ Feb 28, 2016 at 16:39
  • $\begingroup$ @IvanNeretin Uh..... Then why is it not mentioned? Also, if it does react, what about the other possible combinations? Do they all occur? If so, shouldn't the products also react? If we perform an experiment in a lab, shouldn't that throw off everything?(So many questions, I know..) And thank you for the welcome.. :D $\endgroup$ Feb 28, 2016 at 16:44
  • $\begingroup$ @IvanNeretin OK.. Never expected that'd be possible, but thanks. Also, can you post that as an answer? I'd like to mark it as one. $\endgroup$ Feb 28, 2016 at 16:53

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Your example is not quite the case when multiple reactions are possible. Also, $\ce{H+}$ not only can, but actually does combine with $\ce{CO3^2-}$ to form a new particle: $\ce{HCO3-}$, the hydrocarbonate anion. This is where it would end if you'd use just one tiny droplet of $\ce{HCl}$. Otherwise, that anion would attract another $\ce{H+}$ and form carbonic acid $\ce{H2CO3}$, which is unstable and decomposes into $\ce{H2O}$ and $\ce{CO2}$, to produce the summary equation which you already know.

As for your question in general, in most cases when so many reactions are possible, they indeed occur all at once. It just so happens that they go at different rates, and one of them outruns the others by a wide margin, so you may consider it "the only one" and safely ignore the rest. Or it does not, and you end up with a bucketful of brown goo. Such examples rarely make it to the textbooks, though.

All this is especially big deal in the field of organic chemistry, where all reactions produce more or less "side products". Read the details on just about any reaction, and you'll see these yield figures to the right: 96%, 92%... or maybe less if the authors were less lucky. That's how much of our reactants went down the desired pathway. The rest went someplace else.

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    $\begingroup$ Well, I don't think this quite explains the gist. // Regardless of pH, there is an equilibrium between $\ce{H+}$, $\ce{CO3^2-}$, $\ce{HCO3-}$, $\ce{H2CO3(aq)}$, $\ce{CO2(aq)}$, and $\ce{CO2(g)}$ where $\ce{CO2(g)}$ is the concentration of $\ce{CO2}$ in the atmosphere. The point is that the atmosphere is such a large sink for $\ce{CO2}$ that the reaction is driven towards pushing the carbonate out of solution. So mass balance wise 99.9+% of the carbonate is converted to $\ce{CO2}$ in the atmosphere. But there is still some minuscule amount of the other carbonate species in solution. $\endgroup$
    – MaxW
    Feb 28, 2016 at 17:48
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    $\begingroup$ OK, point taken. So even in a reaction that simple, the yield is not quite 100%. Wait, it is never 100%, because the equilibrium constant is never $\infty$. $\endgroup$ Feb 28, 2016 at 19:48

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