The reaction of gold(III) sulfate (Au₂(SO₄)₃) with barium chloride (BaCl₂) results in a double displacement reaction where the cations and anions switch partners. This leads to the formation of gold(III) chloride (AuCl₃) and barium sulfate (BaSO₄).

Here's the balanced chemical equation:

Au₂(SO₄)₃ (aq) + 3BaCl₂ (aq) → 2AuCl₃ (aq) + 3BaSO₄ (s)

Explanation:

* Gold(III) sulfate (Au₂(SO₄)₃) is a soluble ionic compound, meaning it dissolves in water to form Au³⁺ and SO₄²⁻ ions.

* Barium chloride (BaCl₂) is also a soluble ionic compound, dissolving in water to form Ba²⁺ and Cl⁻ ions.

* When the solutions are mixed, the Au³⁺ ions react with the Cl⁻ ions to form gold(III) chloride (AuCl₃), which is also soluble.

* The Ba²⁺ ions react with the SO₄²⁻ ions to form barium sulfate (BaSO₄), which is insoluble in water.

Key Observations:

* A white precipitate of barium sulfate will form, indicating the completion of the reaction.

* The solution will remain yellow due to the presence of gold(III) chloride.

Note: This reaction is often used to precipitate barium sulfate, which is used in various applications like medical imaging and manufacturing.