Question

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

Short answer

A is \( \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \), B is \( \mathrm{Al}_{2} \mathrm{O}_{3} \), C is \( \mathrm{SO}_{3} \).

Step-by-step solution

Write down the chemical reaction

Start with the given reaction: \( \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\cdot 18 \mathrm{H}_{2} \mathrm{O} \xrightarrow{\text{Heat}} -18 \mathrm{H}_{2} \mathrm{O} + \mathrm{A} \xrightarrow{800^{\circ} \mathrm{C}} \mathrm{B} + \mathrm{C} \).

Key concepts

Dehydration of Hydrates

Hydrates are compounds that have water molecules weakly bonded to them. When these substances are heated, they lose the water, resulting in a process known as dehydration. In our exercise, the compound \[ \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\cdot 18 \mathrm{H}_{2} \mathrm{O} \] initially contains 18 water molecules per formula unit. As the temperature increases, the heat causes these water molecules to evaporate, transforming the hydrate into an anhydrous form, which means it no longer contains water. Key Points About Dehydration:

• The heat breaks the bonds between the water molecules and the salt compound.
• This process is essential for determining the anhydrous state of a hydrated compound.
• It typically involves several steps, with water gradually being released as the temperature rises.
• The end result for our compound is the anhydrous form of the sulfate, \[ \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \].

Dehydration plays a crucial role in understanding chemical reactions involving hydrates, as it is often the first step in such processes.

Thermal Decomposition

Thermal decomposition is the process where a chemical compound breaks down into its constituent substances when heated to a certain temperature. This is what happens to \( \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \) after it loses its water molecules from dehydration.Understanding the Reaction:

• Once the aluminum sulfate loses its water, the anhydrous compound decomposes further upon heating to a high temperature.
• In our case, this occurs at around 800°C, where \( \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \) decomposes into aluminum oxide (\( \mathrm{Al}_{2}\mathrm{O}_{3} \)) and sulfur trioxide (\( \mathrm{SO}_{3} \)).
• Thermal decomposition involves breaking strong chemical bonds, often requiring significant energy.
• This type of reaction is typical in inorganic chemistry and is critical for understanding how compounds behave under heat.

Thermal decomposition is a useful concept in fields ranging from materials science to environmental studies, helping us to predict how substances will behave when subjected to high temperatures.

Inorganic Chemistry

Inorganic chemistry focuses on compounds that do not primarily consist of carbon-hydrogen bonds. It encompasses a diverse array of elements including metals, minerals, and organometallics. Our focus in this exercise is on aluminum sulfate \( \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \), which is a prominent subject of investigation in inorganic chemistry.Significant Aspects:

• Inorganic compounds often have simple structures and are typically categorized by the ions they form.
• Reactions like this one provide insight into the properties of metals and their compounds.
• Aluminum sulfate, for instance, is widely used in water treatment processes and as a mordant in dyeing.
• Keen understanding of the thermal behavior of inorganic compounds is essential in many industrial processes.

Inorganic chemistry is essential for understanding the myriad of materials that make up the world beyond organic carbon compounds, emphasizing reactions like dehydration and decomposition to demonstrate their varied applications.