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They seem all correct to me, is there any reason why one reaction would happen over the other(s)?

$$\begin{align} \ce{K2CO3 + 2 HCl &-> 2 KCl + H2CO3}\\[6pt] \ce{K2CO3 + 2 HCl &-> 2 KCl + H2O + CO2}\\[6pt] \ce{K2CO3 + HCl &-> KHCO3 + H2O} \end{align}$$

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  • $\begingroup$ Welcome to Chemistry.SE! Feel free to take the tour to learn more about this site. Your question appears to be a homework question, and in its current form doesn't comply with our homework policy, and is subject to removal. Please edit your question accordingly. Visit this link to learn about MathJax and mhchem $\endgroup$ – getafix Aug 29 '16 at 2:19
  • $\begingroup$ @getafix Was not a homework question, but I've edited it accordingly anyway, thanks. $\endgroup$ – Anonymous Aug 29 '16 at 2:33
  • $\begingroup$ If you read the policy, a "homework question", doesn't have to be from an actual homework assignment. You must demonstrate some effort towards solving the problem, or at least share your thoughts with us on the matter. $\endgroup$ – getafix Aug 29 '16 at 2:36
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They are all correct and they will all stay in equilibrium with one another, whereas the products of the second equation will be the most favoured, since carbonic acid is an unstable weak acid and will likely dissociate into $\ce{CO_2}$ and $\ce{H_2O}$. At the same time potassium hydrogen carbonate will get protonated easily: $$\ce{HCl + KHCO_3 \rightarrow KCl + H_2CO_3}$$ again resulting in more carbon dioxide and water.

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  • $\begingroup$ The real point about equation two is that this reaction is typically taking place in aqueous solution, and the $\ce{CO2}$ escapes from liquid into gas phase. $\endgroup$ – MaxW Aug 30 '16 at 5:00
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All the equations are correct. The first two are basically the same. Carbonic acid is a weak acid. It is less stable. So it dissociates itself to give $\ce{CO2}$ and $\ce{H2O}$. And the last one is just based on the simple difference in the moles. So you can use the equation depending on the moles.

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  • 1
    $\begingroup$ Could you please fix that caps lock first? $\endgroup$ – M.A.R. ಠ_ಠ Aug 29 '16 at 17:57

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