91Ó°ÊÓ

Molecular equations give a full picture of the chemical reaction by showing all the reactants and products in their molecular forms. In the given exercise, the first molecular equation is:
$$ \text{Ba(NO}_3\text{)}_{2(aq)} + \text{Na}_{2}\text{CO}_3\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{NaNO}_3\text{(aq)} $$
Think of these equations as the big snapshot of what's happening. All compounds are represented before they dissociate in water. For the exercise, other valid sets of molecular equations would be:
$$ \text{BaCl}_2\text{(aq)} + \text{K}_2\text{CO}_3\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{KCl}\text{(aq)} $$
$$ \text{Ba(CH}_3\text{CO}_2\text{)}_{2(aq)} + \text{(NH}_4\text{)}_2\text{CO}_3\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{NH}_4\text{CH}_3\text{CO}_2\text{(aq)} $$
Notice that these equations fully display the reactants and products without specifying ionic forms.

Total ionic equations break down all aqueous substances into their respective ions. This provides a more granular look at the reaction. Based on the exercise:
The total ionic equation for the reaction:
$$ \text{Ba(NO}_3\text{)}_{2(aq)} + \text{Na}_2\text{CO}_3\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{NaNO}_3\text{(aq)} $$
becomes:
$$ \text{Ba}^{2+}\text{(aq)} + 2\text{NO}_3{}\text{(aq)} + 2\text{Na}^{+}\text{(aq)} + \text{CO}_3{}^{2-}\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{Na}^{+}\text{(aq)} + 2\text{NO}_3{}\text{(aq)} $$
In this step, we express all reactants and products that are strong electrolytes in their dissociated forms. Similarly, breaking down other molecular equations:
For: $$ \text{BaCl}_2\text{(aq)} + \text{K}_2\text{CO}_3\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{KCl}\text{(aq)} $$
we get:
$$ \text{Ba}^{2+}\text{(aq)} + 2\text{Cl}^{-}\text{(aq)} + 2\text{K}^{+}\text{(aq)} + \text{CO}_3{}^{2-}\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{K}^{+}\text{(aq)} + 2\text{Cl}^{-}\text{(aq)} $$
Remove spectator ions to reveal the net ionic equation next.

A chemical reaction involves the rearrangement of atoms to form new substances. The molecular equations mentioned above show this at a macroscopic level. The total ionic and net ionic equations zoom in to depict the exact particles involved.
In a reaction like:
$$ \text{BaCl}_2\text{(aq)} + \text{K}_2\text{CO}_3{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{KCl}\text{(aq)} $$
Barium ions (\text{Ba}^{2+}) react with carbonate ions (\text{CO}_3{}^{2-}) to form the solid barium carbonate (\text{BaCO}_3), indicating a precipitation reaction. This shows how ions in solution form a solid.
Chemical reactions can be varied, such as acid-base reactions, redox reactions, and precipitation reactions like our example here. They form the core of much of chemistry learning and applications.

Spectator ions are ions in the solution that do not participate in the formation of the precipitate or the reaction, essentially 'watching' the other ions react. For example, in the total ionic equation:
$$ \text{Ba}^{2+}\text{(aq)} + 2\text{Cl}^{-}\text{(aq)} + 2\text{K}^{+}\text{(aq)} + \text{CO}_3{}^{2-}\text{(aq)} \rightarrow \text{BaCO}_3\text{(s)} + 2\text{K}^{+}\text{(aq)} + 2\text{Cl}^{-}\text{(aq)} $$
$$ 2\text{Cl}^{-} $$ and $$ 2\text{K}^{+} $$ are spectator ions because they appear unchanged on both sides of the equation. They do not affect the formation of the precipitate (the solid product) and can be omitted to simplify the net ionic equation. Removing these spectator ions highlights the actual reacting species and the real transformation taking place during the reaction.
Identifying and removing spectator ions simplifies the understanding of chemical reactions, focusing on the core chemical change.