Problem 23 - A Complete Resource Book in Chemistry for JEE Main

Question

The product obtained on fusion of \(\mathrm{BaSO}_{4}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) is (a) \(\mathrm{BaO}\) (b) \(\mathrm{BaCO}_{3}\) (c) \(\mathrm{BaHS}_{4}\) (d) \(\mathrm{Ba}(\mathrm{OH})_{2}\)

Step-by-Step Solution

Answer

The product is \( \mathrm{BaCO}_{3} \).

Step 1: Write the chemical equation

To identify the product of the fusion of \( \mathrm{BaSO}_{4} \) and \( \mathrm{Na}_{2} \mathrm{CO}_{3} \), we start by writing the chemical reaction equation. When \( \mathrm{BaSO}_{4} \) is heated with \( \mathrm{Na}_{2} \mathrm{CO}_{3} \), the products typically include \( \mathrm{BaCO}_{3} \) and \( \mathrm{Na}_{2} \mathrm{SO}_{4} \). Therefore, the equation is: \[ \mathrm{BaSO}_{4} + \mathrm{Na}_{2} \mathrm{CO}_{3} \rightarrow \mathrm{BaCO}_{3} + \mathrm{Na}_{2} \mathrm{SO}_{4} \]

Step 2: Analyze the products

From the equation in Step 1, we see that the main barium-containing product formed is \( \mathrm{BaCO}_{3} \). This happens because \( \mathrm{Ba}^{2+} \) reacts with \( \mathrm{CO}_{3}^{2-} \) ions to form \( \mathrm{BaCO}_{3} \), while \( \mathrm{Na}^{+} \) and \( \mathrm{SO}_{4}^{2-} \) are left to form \( \mathrm{Na}_{2} \mathrm{SO}_{4} \).

Key Concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances called products. This involves breaking chemical bonds and forming new ones. The exercise provided is an example of a chemical reaction where two reactants, barium sulfate \( (\mathrm{BaSO}_{4}) \) and sodium carbonate \( (\mathrm{Na}_{2}\mathrm{CO}_{3}) \), combine to form new products.
The reaction can be represented as:

• Barium sulfate \( + \) Sodium carbonate \( \rightarrow \) Barium carbonate \( + \) Sodium sulfate

This process involves displacement and exchange of ions. The sulphate ions (\(\mathrm{SO}_{4}^{2-}\)) from barium sulfate engage with sodium ions (\(\mathrm{Na}^{+}\)), forming sodium sulfate, while the carbonate ions \((\mathrm{CO}_{3}^{2-})\) from sodium carbonate react with barium ions \((\mathrm{Ba}^{2+})\) to form barium carbonate.

Stoichiometry

Stoichiometry is the area of chemistry that deals with the quantitative relationships between elements and compounds in a chemical reaction. Here, understanding stoichiometry helps us balance the chemical equation correctly. The equation \[\mathrm{BaSO}_{4} + \mathrm{Na}_{2}\mathrm{CO}_{3} \rightarrow \mathrm{BaCO}_{3} + \mathrm{Na}_{2}\mathrm{SO}_{4}\] is already balanced, showing an equal number of atoms for each element on both sides.
Through stoichiometry, you can calculate the moles and weights of reactants and products based on their chemical formulas.

• The molar mass of the reactants and products is crucial for this calculation.
• This balance ensures that matter is conserved during the reaction.

Stoichiometry is key for predicting how much product can be formed from given amounts of reactants, which is essential in planning reactions and scaling them for industrial processes.

Barium Compounds

Barium compounds are interesting due to their distinct chemical properties. Barium carbonate \((\mathrm{BaCO}_{3})\), the product in our reaction, is one such compound with resounding industrial significance.

• Barium carbonate appears as a white powder and is sparingly soluble in water, which can limit its reactivity in some scenarios.
• It is often used in ceramics, bricks, and rat poison.
• Its formation in the given chemical reaction is pivotal, as the conversion of \(\mathrm{BaSO}_{4}\) to \(\mathrm{BaCO}_{3}\) showcases its ability to form from interaction with carbonate ions.

Understanding barium compounds aids in predicting their use and behavior in various chemical environments. Furthermore, knowing their reactivity and solubility assists in safety and handling when used in laboratory or industrial applications.