In this example, we will write the molecular, complete ionic, and net ionic equations for a precipitation reaction. Our problem reads: "Write the balanced molecular, complete ionic, and net ionic equations for the reaction between aqueous aluminum sulfate and aqueous calcium nitrate." We'll start by writing the molecular equation. This is going to be a precipitation reaction. Aqueous aluminum sulfate will react with calcium nitrate to produce aqueous aluminum nitrate and solid calcium sulfate. In this precipitation reaction, when we switch the cations and anions in these ionic compounds, calcium sulfate is insoluble, so that is the precipitate. To balance the molecular equation, we can balance the ions. Right now, on the left, we have two aluminum three-plus ions in aluminum sulfate. So, to get two aluminum on the right, we need two aluminum nitrates. We also have three sulfates on the left with one aluminum sulfate. So, to get three sulfates on the right, we need three calcium sulfates. We currently have three calcium two-plus on the right, and to get three calcium two-plus on the left, we'll need three calcium nitrates. We can double-check to make sure we have the same number of nitrates on both sides, and we do have six nitrates on both sides. So, this should be our balanced molecular equation. To write the complete ionic equation, we're going to break apart any strong electrolytes. A strong electrolyte is a soluble salt or a strong acid. In this case, we just have soluble salts to break apart. Since aluminum sulfate is a soluble salt, it shows up as aqueous in our reaction. And our coefficient is 1. The subscript of two aluminum in aluminum sulfate means we have two aluminum three-plus ions. For sulfates, we have three sulfates in aluminum sulfate, so we end up with three sulfate ions. Calcium nitrate is also aqueous, so we'll split it apart since it's a strong electrolyte. As a soluble salt, we have three calcium two-plus ions, and we have three times two nitrates, so six nitrate ions. For our products, we're producing aluminum nitrate, which is also a strong electrolyte as a soluble salt. So, we'll break it apart into aluminum three-plus ions and six nitrate ions. Our calcium sulfate is an insoluble salt, which means it's not a strong electrolyte. So, we do not break it apart in the complete ionic equation. It has to remain a solid. For our net ionic equation, we have to eliminate spectator ions. Spectator ions show up on both sides of the equation. In the complete ionic equation, there are two aluminum three-plus ions on both sides, and there are six nitrate ions on both sides. So, we can eliminate these ions, and what is left is our net ionic equation. Therefore, the net ionic equation is: three sulfate plus two calcium two-plus produces three calcium sulfate solid. These are our molecular, complete ionic, and net ionic equations.