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

Question

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

The products A, B, and C are respectively (a) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{3}\).

Step 1: Understanding the Chemical Reaction

We start with the compound \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 18 \mathrm{H}_{2} \mathrm{O}\). It undergoes a decomposition reaction when heat is applied, resulting in the loss of 18 water molecules as steam \(-18 \mathrm{H}_{2} \mathrm{O}\). This process leads to the formation of an anhydrous compound \(A\). At a high temperature of 800°C, further decomposition occurs, resulting in compounds \(B\) and \(C\).

Step 2: Identify Compound A

The initial process of heating \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 18 \mathrm{H}_{2} \mathrm{O}\) leads to the loss of all water molecules, resulting in anhydrous \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). Thus, compound \(A\) is \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\).

Step 3: Identify Compounds B and C

Upon further heating at 800°C, \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) decomposes to form \(\mathrm{Al}_{2} \mathrm{O}_{3}\) and \(\mathrm{SO}_{3}\). Thus, compounds \(B\) and \(C\) are \(\mathrm{Al}_{2} \mathrm{O}_{3}\) and \(\mathrm{SO}_{3}\) respectively.

Step 4: Select the Correct Answer

From our analysis, \(A\) is \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), \(B\) is \(\mathrm{Al}_{2} \mathrm{O}_{3}\), and \(C\) is \(\mathrm{SO}_{3}\). Compare these with the given options: (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) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{2}\). The correct answer is (a).

Key Concepts

Chemical Equation Balancing

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is followed. In every chemical reaction, the same number and type of atoms must exist before and after the reaction. This means that the mass of reactants must equal the mass of products. It keeps everything balanced, much like a scale.

When you look at a chemical equation, it's important to count the atoms of each element involved and ensure they match on both sides of the reaction. This often requires adjusting the coefficients, which are the numbers placed before the chemical formulas. Here are some tips to make balancing easier:

• Start by balancing the most complex molecule first, as it often contains the highest number of atoms.
• Leave hydrogen and oxygen for last, as they frequently appear in multiple compounds.
• Use fractional coefficients initially if necessary, and multiply through by the denominator to clear fractions when finished.

In our specific exercise, balancing isn't explicitly explored because the compounds involved decompose. However, even in decomposition, ensuring the same number of atoms on both sides can guide understanding and accuracy.

Hydration and Dehydration

Hydration and dehydration reactions are notable processes in chemistry. They involve the addition or removal of water molecules from a compound, altering its physical and chemical properties.

**Hydration**
This is when water molecules are added. For example, many metal salts form crystals that include water molecules, known as hydrates. These water molecules can often be removed by applying heat.

**Dehydration**
Dehydration, on the other hand, is the opposite process where water is removed. This is what happens in our exercise. When aluminum sulfate hydrate, \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3} \cdot 18 \mathrm{H}_{2} \mathrm{O}\), is heated, it loses all 18 water molecules, resulting in an anhydrous form.

Understanding these processes is crucial because they can significantly alter a compound's structure and reactivity, affecting how substances are used and behave in various applications and reactions.

Aluminum Sulfate

Aluminum sulfate, \(\mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}\), is a chemical compound commonly used in water purification and waste treatment as a coagulant. It helps to remove impurities by causing them to clump together and settle out of water.

**Composition and Properties**

• It includes aluminum, sulfur, and oxygen atoms.
• Has a strong ability to attract and bind with other particles, making it effective in clarifying water.

In our exercise, aluminum sulfate undergoes a dehydration process which showcases its potential to undergo changes upon heating.

**Reactions Involving Aluminum Sulfate**
Upon heating, aluminum sulfate decomposes into its constituent parts: \(\mathrm{Al}_{2}\mathrm{O}_{3}\) and \(\mathrm{SO}_{3}\). This reaction exemplifies a decomposition reaction where a single compound breaks down into two or more different substances, often due to the application of heat.

Understanding such characteristics of aluminum sulfate is essential for its application in industrial processes and environmental management.