If you mixed solid sodium carbonate with HCl (aq) the sodium carbonate dissolves in the water solution of acid, becoming (aq).
Now if you use CaCO3 instead, it is not soluble in the water, so what happens is the acid attacks the ionic structure of the chalk, breaking it apart. Yes, I would put CaCO3 (s) on the left side.
I am rather confused about writing ionic equations.
1) SOLID sodium carbonate reacts with hydrochloric acid to give NaCl,CO2 & H2O.
Is the ionic equation
CO32-(aq)+2H+(aq)-> CO2(g) + H2O(l)?
Do I consider SOLID sodium carbonate to be in aqueous state since it is a soluble ionic compound? Or do I have to write Na2CO3(s) on the left hand side of the equation?
2) If I use CaCO3 instead, I will have to write CaCO3(s) on the left hand side of the ionic equation?
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and an additional note: By adding the CaCO3 (s) on the left, you have to add Ca+2 on the right to balance charge.
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