Problem 93 - A Complete Resource Book in Chemistry for JEE Main

Question

In the reaction: \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 18 \mathrm{H}_{2} \mathrm{O} \frac{\mathrm{Heat}}{-18 \mathrm{H}_{2} \mathrm{O}}-\mathrm{A} \stackrel{800^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{B}+\mathrm{C}\) The product \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are respetively (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) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{2}\)

Step-by-Step Solution

Answer

Option (a): \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}, \mathrm{Al}_{2}\mathrm{O}_{3}, \mathrm{SO}_{3}\).

Step 1: Understanding the given reaction

The equation provided involves heating a hydrated form of aluminum sulfate: \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 18 \mathrm{H}_{2}\mathrm{O}\) to remove the water molecules and decompose the compound at a specified temperature to form products A, B, and C.

Step 2: Dehydration process

First, the water molecules in \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} \cdot 18 \mathrm{H}_{2}\mathrm{O}\) are removed upon heating, yielding \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\) as product A.

Step 3: Thermal decomposition

Heating \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\) at a high temperature (\(800^{\circ} \mathrm{C}\)) causes it to decompose into aluminium oxide \(\mathrm{Al}_{2}\mathrm{O}_{3}\) (product B) and sulfur trioxide \(\mathrm{SO}_{3}\) (product C).

Step 4: Identifying correct option

From the process, products A, B, and C are identified as \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\), \(\mathrm{Al}_{2}\mathrm{O}_{3}\), and \(\mathrm{SO}_{3}\) respectively. The correct option matching this outcome is (a).

Key Concepts

Aluminum Sulfate

Aluminum sulfate is a chemical compound with the formula \( \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} \). It is most commonly found in a hydrated form, which means it combines with water molecules, as seen in the reaction \( \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} \cdot 18 \mathrm{H}_{2} \mathrm{O} \). This means that for every molecule of aluminum sulfate, there are 18 water molecules attached to it.
This compound is widely used in water purification and paper manufacturing.

• In water treatment, it helps to coagulate impurities, making them easier to remove from water.
• In papermaking, it plays a role in sizing the paper, affecting its ability to absorb ink.

However, when heated, as in the reaction above, aluminum sulfate behaves differently. It undergoes decomposition, releasing its attached water molecules and breaking down into other substances. Understanding its behavior under heat is crucial for industrial applications where thermal stability is important.

Dehydration Process

The dehydration process involves the removal of water molecules from a hydrated compound through heating. In the current reaction, when \( \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} \cdot 18 \mathrm{H}_{2} \mathrm{O} \) is heated, it loses its #-water molecules, becoming simply \( \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} \).
This process is important because:

• It transforms the compound from a hydrated to an anhydrous state (meaning without water).
• It prepares the compound for further reactions, such as decomposition.

Dehydration isn't just about losing water; it often results in significant changes in the properties of a compound. Anhydrous compounds can have different melting points, densities, and reactivity compared to their hydrated forms. In industrial processes, controlling the phase of dehydration is crucial for obtaining a specific outcome in product formation.

High Temperature Reactions

High temperature reactions involve chemical changes that occur under elevated temperatures. In the case of aluminum sulfate, it decomposes when heated to a temperature of \(800^{\circ} \mathrm{C}\).
Here's what happens at these high temperatures:

• **Decomposition:** The compound breaks down into aluminum oxide \(\mathrm{Al}_{2}\mathrm{O}_{3}\) and sulfur trioxide \(\mathrm{SO}_{3}\). These are stable compounds that result from the breakdown of the sulfate ion and aluminum.
• **Application:** Such decomposition reactions are vital in industrial settings where metals are extracted and refined.

Reactions at high temperatures often result in the creation of new materials with different properties. This is essential in industries concerned with materials engineering or chemical manufactury. Understanding these reactions allows chemists to harness heat efficiently to drive chemical processes to desirable ends, transforming raw materials into useful products.