Question

In the reaction: \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} .18 \mathrm{H}_{2} \mathrm{O} \frac{\text { heat }}{-18 \mathrm{H}_{2} \mathrm{O}}\) \(\mathrm{A} \quad{ }_{90^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{B}+\mathrm{C}\). The product \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are respectively (a) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{3}\) (b) \(\mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}, \mathrm{SO}_{3}\) (c) \(\mathrm{Al}_{2} \mathrm{SO}_{4}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{3}\) (d) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{2}\)

Short answer

The correct answer is (a) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{3} \).

Step-by-step solution

Understanding the Reaction

The chemical reaction involves the heating of hydrated aluminum sulfate, \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} .18 \mathrm{H}_{2} \mathrm{O}\), to remove the water and produce anhydrous sulfate and secondary products. The equation given is \( \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} .18 \mathrm{H}_{2} \mathrm{O} \xrightarrow{90^{\circ} \mathrm{C}} \text{--} 18 \mathrm{H}_{2} \mathrm{O} + \mathrm{B} + \mathrm{C} \). Our goal is to identify \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\).

Determine Substance A

Upon heating, \( \mathrm{A} \) is what remains of the hydrated compound after losing water. \( \mathrm{A} \) is the anhydrous form of the original compound: \( \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \).

Identify Products B and C

Upon further heating, the anhydrous aluminum sulfate \( \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \) decomposes into aluminum oxide \( \mathrm{Al}_{2} \mathrm{O}_{3} \) and sulfur trioxide \( \mathrm{SO}_{3} \). \( \mathrm{B} \) is \( \mathrm{Al}_{2} \mathrm{O}_{3} \) and \( \mathrm{C} \) is \( \mathrm{SO}_{3} \).

Verify with Options

Checking the provided options: (a) shows the correct order for \( \mathrm{A} \), \( \mathrm{B} \), and \( \mathrm{C} \) as \( \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{3} \). This matches our findings.

Key concepts

Thermal Decomposition

Thermal decomposition is a type of chemical reaction where a compound is broken down into simpler substances when heated. This process involves absorbing heat energy to overcome the bonds holding the compound's atoms together. During heating, the compound undergoes a physical change, resulting in the production of different chemical substances. In the case of aluminum sulfate - real example is when hydrated aluminum sulfate is heated, the water molecules trapped in the crystal structure are released, and further decomposition results in the formation of an anhydrous compound and other byproducts. - The thermal decomposition of aluminum sulfate leads to the creation of aluminum oxide and sulfur trioxide, demonstrating how heat can cause complex compounds to disassemble into simpler products. - Such reactions are commonly used in industrial processes for material transformation and extraction of chemicals.

Hydrated Compounds

Hydrated compounds are substances that contain water molecules within their crystal structure. These water molecules, often referred to as "water of crystallization," are integral to the compound's structure and properties. For example, - hydrated aluminum sulfate \ \( \text{Al}_2(\text{SO}_4)_3 \cdot 18 \text{H}_2\text{O} \) is a complex where water is present in the compound's framework. - The presence of these water molecules affects the compound's weight, volume, and sometimes even its color. - When subjected to heat, hydrated compounds release the incorporated water molecules, transitioning into an anhydrous state. - This process alters the compound's physical state and properties, such as in the transformation of hydrated aluminum sulfate by removing its water content at 90 °C, preparing it for further reactions.

Anhydrous Compounds

Anhydrous compounds are substances that either naturally do not contain water or have had their water content removed. - Anhydrous substances are often the result of heating hydrated compounds to evaporate the water of crystallization, achieving a new chemical structure without water.- For example, heating hydrated aluminum sulfate converts it into its anhydrous form, \ \( \text{Al}_2(\text{SO}_4)_3 \). - Anhydrous compounds exhibit different chemical properties compared to their hydrated counterparts and are essential in various applications, ranging from drying agents to catalysts in industrial processes. - The transition from hydrated to anhydrous forms is integral in processes like chemical analysis and synthesis, altering how materials react chemically.

Sulfur Trioxide Formation

Sulfur trioxide \( \text{SO}_3 \) is a significant chemical in industrial applications, primarily known for its role in producing sulfuric acid. When heating aluminum sulfate, one of the key reactions is the release of sulfur trioxide as a byproduct. - This process showcases the potential of thermal decomposition to generate commercially valuable chemicals. - The gradual decomposition of the compound causes sulfur-containing sections to release as sulfur trioxide gas, which can be captured or further processed. - Sulfur trioxide's formation is critical due to its reactivity and utility in creating acids, demonstrating the chemical transformation capabilities when applying heat to complex compounds. - Understanding its production through decomposition helps in appreciating the resourceful nature of chemical reactions, where multiple useful products can be derived from a single compound.