Respuesta :
Answer: Balanced molecular equation :
[tex]2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)[/tex]
Total ionic equation:
[tex]6Na^+(aq)+3PO_4^{2-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)[/tex]
The net ionic equation:
[tex]2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)[/tex]
Explanation:
Complete ionic equation : In complete ionic equation, all the substances that are strong electrolyte are present in an aqueous state as ions.
Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
When sodium phosphate and zinc acetate then it gives zinc phosphate and sodium acetate as product.
The balanced molecular equation will be,
[tex]2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)[/tex]
The total ionic equation in separated aqueous solution will be,
[tex]6Na^+(aq)+2PO_4^{3-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)[/tex]
In this equation, and are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)[/tex]
