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Write the ionic reaction equation for the reaction

\[\ce{Ca(OH)2(aq) + CO2(g) <=> CaCO3(s) + H2O(l)}.\]

I first wrote the complete ionic equation as such: \[\ce{Ca^2+ + 2 OH^- + CO2 <=> Ca^2+ + CO3^2- + H2O}.\tag{R1}\] Then I crossed out the common $\ce{Ca^2+}$, and got \[\ce{2OH^- + CO2 <=> CO3^2- + H2O}.\tag{R2}\] How is this wrong?

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    $\begingroup$ Did you consider solubility rules? $\endgroup$
    – Mathew Mahindaratne
    Commented Feb 24, 2023 at 16:40
  • $\begingroup$ Also, some CO2 might be considered carbonic acid, en.wikipedia.org/wiki/Carbonic_acid $\endgroup$
    – DrMoishe Pippik
    Commented Feb 24, 2023 at 17:47
  • $\begingroup$ It's a trick question. There aren't any spectator ions you could cancel out. In fact, unless you write it in steps like in Maurice's answer, there are only two ions (from the soluble calcium hydroxide). I will edit your question to add the physical states, that might clear things up. Carbon dioxide could also be written as in aqueous solution, that is a matter of preference or how the reaction is set up. $\endgroup$
    – Karsten ♦
    Commented Feb 24, 2023 at 19:13

1 Answer 1

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$\ce{CaCO3}$ is a precipitate. It is not made of separated $\ce{Ca^{2+}}$ and $\ce{CO3^{2-}}$ ions. So the equation is : $$\ce{Ca^{2+} + 2 OH^- + CO2 -> CaCO3 + H2O}$$ This equation may be considered as two successive equations, namely :$$\ce{CO2 + 2 OH^- -> CO3^{2-} + H2O}$$ and $$\ce{Ca^{2+} + CO3^{2-} -> CaCO3}$$

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