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?

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.

and an additional note: By adding the CaCO3 (s) on the left, you have to add Ca+2 on the right to balance charge.

1) When writing an ionic equation, it is important to consider the state of each species involved in the reaction. However, it's also important to note that not all ionic compounds immediately dissolve in water. In this case, sodium carbonate (Na2CO3) is a soluble compound, so it will typically be written as Na2CO3(aq) on the left-hand side of the equation.

The balanced molecular equation for the reaction between solid sodium carbonate and hydrochloric acid is:
Na2CO3(s) + 2HCl(aq) -> 2NaCl(aq) + CO2(g) + H2O(l)

To write the ionic equation, you need to represent each species as ions if they exist in solution. In this case, chloride ions (Cl-) already exist as ions in hydrochloric acid, so they don't need to be further dissociated. However, sodium carbonate will dissociate into its constituent ions in water.
So the ionic equation becomes:
CO32-(aq) + 2H+(aq) -> 2Na+(aq) + 2Cl-(aq) + CO2(g) + H2O(l)

2) If you use calcium carbonate (CaCO3) instead of sodium carbonate, you would indeed need to write CaCO3(s) on the left-hand side of the ionic equation, as the compound is a solid and does not dissociate into ions in water. The rest of the ionic equation would be the same as before, considering the dissociation of hydrochloric acid and the resulting ions.

Keep in mind that when writing ionic equations, it's crucial to correctly represent the states of the substances involved, taking into account their solubility and ability to dissociate into ions.