The reaction of copper(II) chloride (CuCl₂) and sodium phosphate (Na₃PO₄) is a double displacement reaction that produces copper(II) phosphate (Cu₃(PO₄)₂) and sodium chloride (NaCl).

Here's the balanced chemical equation:

3 CuCl₂ (aq) + 2 Na₃PO₄ (aq) → Cu₃(PO₄)₂ (s) + 6 NaCl (aq)

Here's what happens:

* Copper(II) chloride (CuCl₂) and sodium phosphate (Na₃PO₄) are both soluble ionic compounds that dissociate into ions in solution.

* Copper(II) ions (Cu²⁺) from CuCl₂ react with phosphate ions (PO₄³⁻) from Na₃PO₄ to form copper(II) phosphate (Cu₃(PO₄)₂), which is an insoluble solid that precipitates out of the solution.

* Sodium ions (Na⁺) from Na₃PO₄ react with chloride ions (Cl⁻) from CuCl₂ to form sodium chloride (NaCl), which remains dissolved in solution.

Key observations:

* The reaction produces a blue-green precipitate of copper(II) phosphate.

* The reaction is exothermic, meaning it releases heat.

Note: This reaction is often used in the laboratory to demonstrate double displacement reactions and the formation of precipitates.