Final answer:
To write the balanced complete ionic and net ionic equations for each reaction, we start with the balanced molecular equation and then break down the ionic compounds into their constituent ions. We then cancel out the spectator ions to write the net ionic equation.
Step-by-step explanation:
To write the balanced complete ionic and net ionic equations for each reaction, we need to first write the balanced molecular equation. Then, we can break down the ionic compounds into their constituent ions, indicating their charges and states. Finally, we cancel out the spectator ions to write the net ionic equation.
a. HCl(aq) + LiOH(aq) -> H₂O(l) + LiCl(aq)
Molecular Equation: HCl(aq) + LiOH(aq) -> H₂O(l) + LiCl(aq)
Complete Ionic Equation: H⁺(aq) + Cl⁻(aq) + Li⁺(aq) + OH⁻(aq) -> H₂O(l) + Li⁺(aq) + Cl⁻(aq)
Net Ionic Equation: H⁺(aq) + OH⁻(aq) -> H₂O(l)
b. CaS(aq) + CuCl₂(aq) -> CuS(s) + CaCl₂(aq)
Molecular Equation: CaS(aq) + CuCl₂(aq) -> CuS(s) + CaCl₂(aq)
Complete Ionic Equation: Ca²⁺(aq) + S²⁻(aq) + 2Cu²⁺(aq) + 2Cl⁻(aq) -> CuS(s) + Ca²⁺(aq) + 2Cl⁻(aq)
Net Ionic Equation: S²⁻(aq) + 2Cu²⁺(aq) -> CuS(s)
c. NaOH(aq) + HC₂H₂O₂(aq) -> H₂O(l) + NaC₂H₂O₂(aq)
Molecular Equation: NaOH(aq) + HC₂H₂O₂(aq) -> H₂O(l) + NaC₂H₂O₂(aq)
Complete Ionic Equation: Na⁺(aq) + OH⁻(aq) + H⁺(aq) + C₂H₂O₂⁻(aq) -> H₂O(l) + Na⁺(aq) + C₂H₂O₂⁻(aq)
Net Ionic Equation: OH⁻(aq) + H⁺(aq) -> H₂O(l)
d. Na₃PO₄(aq) + NiCl₂(aq) -> Ni₃(PO₄)₂(s) + NaCl(aq)
Molecular Equation: Na₃PO₄(aq) + NiCl₂(aq) -> Ni₃(PO₄)₂(s) + NaCl(aq)
Complete Ionic Equation: 3Na⁺(aq) + PO₄³⁻(aq) + 2Ni²⁺(aq) + 2Cl⁻(aq) -> 2Ni₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)
Net Ionic Equation: 2PO₄³⁻(aq) + 2Ni²⁺(aq) -> 2Ni₃(PO₄)₂(s)