Write the balanced ionic equation for the reaction sodium bicarbonate with sulphuric acid.

To write the balanced ionic equation for the reaction between sodium bicarbonate (NaHCO₃) and sulfuric acid (H₂SO₄), let's first consider the reaction that takes place. Sodium bicarbonate is a weak base, and sulfuric acid is a strong acid. When they react, sodium bicarbonate will neutralize sulfuric acid, producing sodium sulfate (Na₂SO₄), carbon dioxide (CO₂), water (H₂O), and bicarbonate ions (HCO₃⁻).

### Step 1: Write the Balanced Molecular Equation

The unbalanced reaction is:
\[ \text{NaHCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{CO}_2 + \text{H}_2\text{O} \]

To balance the equation, we need to consider the number of sodium (Na), carbon (C), hydrogen (H), sulfur (S), and oxygen (O) atoms on both sides.

1. Sodium (Na): There is 1 Na on the reactants' side and 2 on the products' side. We need to adjust that.
2. Carbon (C): There is 1 C on both sides.
3. Hydrogen (H): There are 2 H on the reactants' side and 4 on the products' side, so we adjust that.
4. Sulfur (S): There is 1 S on both sides.
5. Oxygen (O): This should balance out as well.

The balanced molecular equation is:
\[ 2 \text{NaHCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{CO}_2 + 2 \text{H}_2\text{O} \]

### Step 2: Write the Ionic Equation

Next, we break down the soluble ionic compounds into their constituent ions. The soluble compounds here are sodium bicarbonate and sulfuric acid. Sodium sulfate, carbon dioxide, and water will remain as is.

The dissociation of the reactants is as follows:
- Sodium bicarbonate: \[ \text{NaHCO}_3 \rightarrow \text{Na}^+ + \text{HCO}_3^- \]
- Sulfuric acid: \[ \text{H}_2\text{SO}_4 \rightarrow 2 \text{H}^+ + \text{SO}_4^{2-} \]

### Step 3: Write the Complete Ionic Equation

The complete ionic equation includes all ions present in the reaction:
\[ 2 \text{Na}^+ + 2 \text{HCO}_3^- + 2 \text{H}^+ + \text{SO}_4^{2-} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{CO}_2 + 2 \text{H}_2\text{O} \]

### Step 4: Identify Spectator Ions

In this case, there are no spectator ions because all ions participate in the reaction.

### Step 5: Write the Balanced Ionic Equation

Since there are no spectator ions, the balanced ionic equation remains the same as the complete ionic equation:
\[ 2 \text{Na}^+ + 2 \text{HCO}_3^- + 2 \text{H}^+ + \text{SO}_4^{2-} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{CO}_2 + 2 \text{H}_2\text{O} \]

### Final Balanced Ionic Equation:
Thus, the final balanced ionic equation for the reaction between sodium bicarbonate and sulfuric acid is:
\[ 2 \text{Na}^+ + 2 \text{HCO}_3^- + 2 \text{H}^+ + \text{SO}_4^{2-} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{CO}_2 + 2 \text{H}_2\text{O} \]