Answer:
SO4^2-(aq) + Sr^2+(aq) → SrSO4(s)
Explanation:
Step 1: Data given
an aqueous solution of strontium chloride = SrCl2(aq)
an aqueous solution of potassium sulfate = K2SO4(aq)
Step 2: The unbalanced equation
K2SO4(aq) + SrCl2(aq) → KCl(aq) + SrSO4(s)
Step 3: Balancing the equation
K2SO4(aq) + SrCl2(aq) → KCl(aq) + SrSO4(s)
On the left side we have 2x K (in K2SO4), on the right side we have 1x K (in KCl). To balance the amount of K on both sides, We have to multiply KCl (on the right side) by 2.
K2SO4(aq) + SrCl2(aq) → 2KCl(aq) + SrSO4(s)
Step 4: The net ionic equation:
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.
2K+(aq) + SO4^2-(aq) + Sr^2+(aq) + 2Cl-(aq) → 2K+(aq) + 2Cl-(aq) + SrSO4(s)
After canceling those spectator ions in both side, look like this:
SO4^2-(aq) + Sr^2+(aq) → SrSO4(s)