Here's how to write the molecular equation for the reaction between lead(II) acetate and potassium chromate:

1. Identify the reactants and products:

* Reactants:

* Lead(II) acetate: Pb(C₂H₃O₂)₂

* Potassium chromate: K₂CrO₄

* Products:

* Lead(II) chromate: PbCrO₄ (a yellow precipitate)

* Potassium acetate: KC₂H₃O₂

2. Write the unbalanced equation:

Pb(C₂H₃O₂)₂ (aq) + K₂CrO₄ (aq) → PbCrO₄ (s) + KC₂H₃O₂ (aq)

3. Balance the equation:

Pb(C₂H₃O₂)₂ (aq) + K₂CrO₄ (aq) → PbCrO₄ (s) + 2 KC₂H₃O₂ (aq)

Therefore, the balanced molecular equation for the reaction is:

Pb(C₂H₃O₂)₂ (aq) + K₂CrO₄ (aq) → PbCrO₄ (s) + 2 KC₂H₃O₂ (aq)