Problem 139 - A Complete Resource Book in Chemistry for JEE Main

Question

The decomposition of a certain mass of \(\mathrm{CaCO}_{3}\) gave \(11.2 \mathrm{dm}^{3}\) of \(\mathrm{CO}_{2}\) gas at STP. The mass of KOH required to completely neutralize the gas is (a) \(56 \mathrm{~g}\) (b) \(28 \mathrm{~g}\) (c) \(42 \mathrm{~g}\) (d) \(20 \mathrm{~g}\)

Step-by-Step Solution

Answer

28 g of KOH are required. The answer is (b).

Step 1: Write the chemical equations

Write the balanced chemical equation for the decomposition of \( \text{CaCO}_3 \) and the reaction of \( \text{CO}_2 \) with \( \text{KOH} \).\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]C\[ \text{CO}_2 + 2 \text{KOH} \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O} \]

Key Concepts

Chemical Equations

Understanding chemical equations is fundamental to grasping the concepts of stoichiometry. A chemical equation is a symbolic representation of a chemical reaction. It shows how reactants transform into products, maintaining the law of conservation of mass.

In our exercise, the decomposition of \( \text{CaCO}_3 \) (calcium carbonate) is expressed as a balanced chemical equation:

• \( \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \)

This equation indicates that calcium carbonate decomposes into calcium oxide (\( \text{CaO} \)) and carbon dioxide (\( \text{CO}_2 \)). The symbols before the arrow represent the reactants, and after the arrow, we have the products.

To solve stoichiometry problems, equations must be balanced, meaning they have equal numbers of each type of atom on both sides of the equation. This ensures that mass is conserved throughout the reaction.

In the context of our problem, another important equation is the reaction of \( \text{CO}_2 \) with \( \text{KOH} \):

• \( \text{CO}_2 + 2 \text{KOH} \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O} \)

This shows that carbon dioxide reacts with potassium hydroxide, forming potassium carbonate and water. By understanding and balancing these equations, we can accurately determine the quantities of substances involved in chemical processes.

Gas Laws

Gas laws are crucial in stoichiometry especially when dealing with gases' volumes, pressure, and temperature. The exercise mentions that \( \text{CO}_2 \) is at Standard Temperature and Pressure (STP).

STP is defined as a temperature of 0°C (or 273.15K) and a pressure of 1 atmosphere (atm). Under these conditions, 1 mole of any gas occupies 22.4 liters (or 22.4 dm³). This relationship is derived from the Ideal Gas Law:

• \( PV = nRT \)

Where:

• \( P \) is the pressure
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the gas constant
• \( T \) is the temperature

For our problem, knowing the volume of \( \text{CO}_2 \) at STP helps us find the amount in moles. We can then use this information to calculate other substances' amounts involved in the reactions based on stoichiometric ratios indicated by the chemical equations.

Understanding gas laws allows us to predict and quantify the behavior of gases in chemical reactions accurately.

Neutralization Reactions

Neutralization reactions are a type of chemical reaction where an acid and a base react to form water and a salt. In our problem, the reaction between \( \text{CO}_2 \) and \( \text{KOH} \) is a neutralization reaction.

\( \text{CO}_2 \) behaves as a weak acid while \( \text{KOH} \) is a strong base. Their reaction follows the equation:

• \( \text{CO}_2 + 2\text{KOH} \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O} \)

In this scenario, \( \text{CO}_2 \) reacts with \( \text{KOH} \) to form \( \text{K}_2\text{CO}_3 \) (potassium carbonate) and water. The balanced equation shows a 1:2 stoichiometric ratio between \( \text{CO}_2 \) and \( \text{KOH} \), meaning 2 moles of \( \text{KOH} \) are required for 1 mole of \( \text{CO}_2 \).

Neutralization reactions are important in various industries and biological systems, where precise control of pH is necessary. By understanding these reactions, we can calculate how much of a base is needed to completely neutralize a given amount of an acid, crucial for solving stoichiometry problems like the one in our exercise.