Type the complete, balanced molecular equation for the reaction between CuCl2 and Na3PO4 in aqueous solution. Include phase labels and format your equation properly. Type the net ionic equation for the reaction between CuCl2 and Na3PO4 in aqueous solution. Include phase labels and format your equation properly.
The Correct Answer and Explanation is:
Complete Balanced Molecular Equation:
3 CuCl2 (aq)+2 Na3PO4 (aq)→Cu3(PO4)2 (s)+6 NaCl (aq)3 \, \text{CuCl}_2 \, (aq) + 2 \, \text{Na}_3\text{PO}_4 \, (aq) \rightarrow \text{Cu}_3(\text{PO}_4)_2 \, (s) + 6 \, \text{NaCl} \, (aq)3CuCl2(aq)+2Na3PO4(aq)→Cu3(PO4)2(s)+6NaCl(aq)
Net Ionic Equation:
3 Cu2+ (aq)+2 PO43− (aq)→Cu3(PO4)2 (s)3 \, \text{Cu}^{2+} \, (aq) + 2 \, \text{PO}_4^{3-} \, (aq) \rightarrow \text{Cu}_3(\text{PO}_4)_2 \, (s)3Cu2+(aq)+2PO43−(aq)→Cu3(PO4)2(s)
Explanation:
When aqueous solutions of copper(II) chloride (CuCl₂) and sodium phosphate (Na₃PO₄) are mixed, a double displacement reaction, also known as a precipitation reaction, occurs. In aqueous solution, both CuCl₂ and Na₃PO₄ dissociate into their respective ions. Copper(II) chloride dissociates into copper(II) ions (Cu²⁺) and chloride ions (Cl⁻), while sodium phosphate dissociates into sodium ions (Na⁺) and phosphate ions (PO₄³⁻).
The reaction proceeds as the Cu²⁺ ions combine with PO₄³⁻ ions to form copper(II) phosphate (Cu₃(PO₄)₂), which is an insoluble solid that precipitates out of the solution. The sodium ions and chloride ions remain dissolved in the solution and are considered spectator ions because they do not participate directly in the formation of the precipitate.
In the complete molecular equation, all species are shown as compounds. Balancing the equation ensures that the number of atoms of each element and the total charge are equal on both sides of the reaction. The balanced molecular equation shows that three moles of CuCl₂ react with two moles of Na₃PO₄ to produce one mole of Cu₃(PO₄)₂ and six moles of NaCl.
The net ionic equation focuses only on the species that form the precipitate. It eliminates the spectator ions and shows that three copper(II) ions react with two phosphate ions to produce solid copper(II) phosphate. This simplification highlights the essence of the chemical change taking place in the reaction.
