Write the balanced net ionic equation for a solution formed from CuCl2 (aq) and Li3PO4 (aq)

3Cu^2+(aq) + 2PO4^-3(aq) →  Cu3(PO4)2 (s)

Explanation:

Step 1: The unbalanced equation

CuCl2 (aq) + Li3PO4 (aq) →  ...

Step 2: Balancing the equation

CuCl2 (aq) + Li3PO4 (aq) → LiCl(aq) + Cu3(PO4)2(s)

On the left side we have 1 time Cu (in CuCl2), on the right side we have 3x Cu (in Cu3(PO4)2.

To balance the amount of Cu we have to multiply CuCl2 (on the left side) by 3.

3CuCl2 (aq) + Li3PO4 (aq) → LiCl(aq) + Cu3(PO4)2(s)

On the left side we have 3x Li, (in Li3PO4), on the right side we have 1x Li (in LiCl).

To balance the amount of Li on both sides, we have to multiply LiCl (on the right side) by 3. But since the amount of Cl on the left side is 6 and on the right side 1. We have to multiply LiCl by 6 and not only by 3

3CuCl2 (aq) + Li3PO4 (aq) → 6LiCl(aq) + Cu3(PO4)2(s)

On the left side we have 1x PO4 (in Li3PO4), in the right side we have 2x PO4 (inCu3(PO4)2). To balanced the amount of PO4 on both sides, we have to multiply Li3PO4 (on the left) by 2. Now the equation is balanced.

3CuCl2 (aq) + 2Li3PO4 (aq) → 6LiCl(aq) + Cu3(PO4)2(s)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will look like this

3Cu^2+(aq) + 2PO4^-3(aq) →  Cu3(PO4)2 (s)

they both have a y in them