Here's how you can figure that out.
In your case, you're dealing with two ionic compounds, potassium bicarbonate,
The reaction takes place in aqueous solution, so you can expect these compounds to be broken up in cations and anions.
The other three species that take part in this reaction, sulfuric acid,
So, let's take them one by one
- Molecular equation
The balanced chemical equation for this reaction looks like this
- Complete ionic equation
This is where you need to show all the ions that are present in solution. Break up the aforementioned ionic compounds to get
The bicarbonate anion has a very, very small dissociation constant (
- Net ionic equation
This time you need to remove spectator ions, which are ions present on both sides of the equation. In your case, you need to remove the potassium cations and the sulfate anions
This is equivalent to
You can divide all the stoichiometric coefficients by 2 to get