Question

Complete and balance this reaction. Then, write the total ionic and the balanced net ionic equation for the following reaction: K₃PO₄ + BaCl₂ →

a) 3K⁺(aq) + PO₄³⁻(aq) + 3Ba²⁺(aq) + 6Cl⁻(aq) → 3Ba₃(PO₄)₂(s)
b) 2K⁺(aq) + PO₄³⁻(aq) + Ba²⁺(aq) + 2Cl⁻(aq) → Ba₃(PO₄)₂(s)
c) K⁺(aq) + PO₄³⁻(aq) + Ba²⁺(aq) + 2Cl⁻(aq) → Ba₃(PO₄)₂(s)
d) 3K⁺(aq) + PO₄³⁻(aq) + 2Ba²⁺(aq) + 6Cl⁻(aq) → 2Ba₃(PO₄)₂(s)

Answer 1

To complete and balance the reaction K₃PO₄ + BaCl₂, we write the balanced equation as 3K₃PO₄ + 2BaCl₂ → Ba₃(PO₄)₂ + 6KCl. The total ionic equation is 3Ba²⁺ + 2PO₄³⁻ → Ba₃(PO₄)₂, and the net ionic equation is 3Ba²⁺ + 2PO₄³⁻ → Ba₃(PO₄)₂.

Step-by-step explanation:

To complete and balance the reaction K₃PO₄ + BaCl₂, we must ensure that the number of atoms on both sides of the equation is equal. The balanced equation is 3K₃PO₄ + 2BaCl₂ → Ba₃(PO₄)₂ + 6KCl. To write the total ionic equation, we separate the aqueous compounds into their dissociated ions, resulting in 3K⁺ + PO₄³⁻ + 2Ba²⁺ + 6Cl⁻ → Ba₃(PO₄)₂ + 6K⁺ + 6Cl⁻. Finally, to write the net ionic equation, we eliminate the spectator ions (those that appear on both sides), giving us 3Ba²⁺ + 2PO₄³⁻ → Ba₃(PO₄)₂.