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.

Taking into account the definition of net ionic equation, a precipitate of SrSO₄ is formed when When an aqueous solution of strontium chloride is added to an aqueous solution of potassium sulfate, and the net ionic equation is:

Sr⁺²(aq) + SO₄⁻²(aq) → SrSO₄(s)

The net ionic equation is a chemical equation for a reaction that lists only the species that participate in the reaction.

In other words, an ionic equation is a chemical equation where electrolytes in aqueous solution are written as dissociated ions. Usually this is a salt dissolved in water.

The balanced equation will be:

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

where (aq) means aqueous and (s) solid.

Taking into account that:

The total ionic equation in separated aqueous solution will be:

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

A spectator ion is an ion that appears both as a reactant and as a product in an ionic equation. Spectator ions can be either cations (positively charged ions) or anions (negatively charged ions).

When writing a net ionic equation, spectator ions found in the original equation are ignored. Thus, the total ionic reaction is different from the net chemical reaction.

In this case, K⁺ and Cl⁻ are the spectator ions because they appear unchanged in both the product and the reagent. So these ions cancel out by writing the net ionic equation and you get:

Sr⁺²(aq) + SO₄⁻²(aq) → SrSO₄(s)

Since SrSO₄ is an insoluble salt, it will precipitate.

Finally, a precipitate of SrSO₄ is formed when When an aqueous solution of strontium chloride is added to an aqueous solution of potassium sulfate, and the net ionic equation is:

Sr⁺²(aq) + SO₄⁻²(aq) → SrSO₄(s)

Learn more:

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

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.