When an aqueous solution of strontium chloride is added to an aqueous solution of potassium sulfate, a precipitation reaction occurs. Write the balanced net ionic equation of the reaction.

Sr²⁺ (aq) + SO₄⁻² (aq) → SrSO₄ (s)

Explanation:

The reactants of our reaction are:

SrCl₂ and K₂SO₄

The products of the reactios are:

SrSO₄ and KCl

We write the molecular reaction:

SrCl₂ (aq) + K₂SO₄ (aq) → 2KCl (aq) + SrSO₄ (s) ↓

To state the net ionic equation we split the salts by their ions from the aqueous solutions. The solid product stays the same.

Sr²⁺ (aq) + 2Cl⁻ (aq) + 2K⁺ (aq) + SO₄⁻² (aq) → 2K⁺ (aq) + 2Cl⁻ (aq) + SrSO₄ (s)

As the potassium cation and the chloride are repeated, they are spectators ions so they won't react in water. Finally our net ionic equation will be:

Sr²⁺ (aq) + SO₄⁻² (aq) → SrSO₄ (s)

Sulfates from elements of group 2, always produces precipitate

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)