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

The mass of KOH required is 56 g (option a).

Step 1: Understanding the Reaction

First, we need to understand the chemical reactions involved. The decomposition of calcium carbonate (\(\text{CaCO}_3\)) produces carbon dioxide (\(\text{CO}_2\)) as follows:\[\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\]The carbon dioxide is then neutralized by potassium hydroxide (\(\text{KOH}\)), according to the reaction:\[\text{CO}_2 + 2\text{KOH} \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O}\]Hence, 1 mole of \(\text{CO}_2\) requires 2 moles of \(\text{KOH}\) for complete neutralization.

Step 2: Calculate Moles of CO2 Produced

At standard temperature and pressure (STP), 1 mole of a gas occupies 22.4 \(\text{dm}^3\). Given that 11.2 \(\text{dm}^3\) of \(\text{CO}_2\) is produced, we can calculate the moles of \(\text{CO}_2\):\[\text{Moles of CO}_2 = \frac{11.2 \text{ dm}^3}{22.4 \text{ dm}^3/\text{mole}} = 0.5 \text{ moles}\]

Step 3: Calculate Moles of KOH Needed

Since 1 mole of \(\text{CO}_2\) requires 2 moles of \(\text{KOH}\), 0.5 moles of \(\text{CO}_2\) will require:\[\text{Moles of KOH} = 0.5 \text{ moles CO}_2 \times 2 = 1 \text{ mole KOH}\]

Step 4: Calculate Mass of KOH Required

The molar mass of \(\text{KOH}\) is calculated as follows: K (39 g/mol) + O (16 g/mol) + H (1 g/mol) = 56 g/mol. Thus, 1 mole of \(\text{KOH}\) has a mass of 56 grams. Therefore, the mass of \(\text{KOH}\) required to neutralize the \(\text{CO}_2\) gas is:\[\text{Mass of KOH} = 1 \text{ mole} \times 56 \text{ g/mole} = 56 \text{ g}\]

Step 5: Verify Answer Against Options

We have calculated the mass of \(\text{KOH}\) needed as 56 g. Checking this against the given options, the correct choice is (a) 56 g.

Key Concepts

Chemical Reactions

Chemical reactions are processes in which substances, known as reactants, are transformed into different substances, called products. In this case, calcium carbonate (\(\text{CaCO}_3\)) decomposes to form calcium oxide (\(\text{CaO}\)) and carbon dioxide (\(\text{CO}_2\)). The reaction can be represented by the equation: - \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\)
Once the carbon dioxide is released, it reacts with potassium hydroxide (\(\text{KOH}\)). Potassium hydroxide, a basic compound, neutralizes the acidic carbon dioxide through the following reaction: - \(\text{CO}_2 + 2\text{KOH} \rightarrow \text{K}_2\text{CO}_3 + \text{H}_2\text{O}\)
This series of reactions highlights the interaction between an acid and a base, demonstrating the principle of neutralization. Understanding these reactions is crucial as it sets the groundwork for balancing equations, predicting products, and understanding stoichiometry.

Mole Concept

The mole concept is a fundamental element of chemistry that provides a bridge between the mass of a substance and the number of particles it contains. It is based on the Avogadro number, which defines that a mole of any substance contains \(6.022 \times 10^{23}\) entities (atoms, molecules, ions, etc.).
In the provided exercise, we calculate moles of carbon dioxide using its volume at standard temperature and pressure (STP). At STP, one mole of any gas occupies \(22.4\ \text{dm}^3\). Since \(11.2\ \text{dm}^3\) of \(\text{CO}_2\) is produced, we determine the moles as follows:- \(\text{Moles of CO}_2 = \frac{11.2 \text{ dm}^3}{22.4 \text{ dm}^3/\text{mole}} = 0.5 \text{ moles}\)
By utilizing the mole concept, we find that \(0.5\) moles of carbon dioxide interact with \(1\) mole of potassium hydroxide since each mole of \(\text{CO}_2\) needs \(2\) moles of \(\text{KOH}\) for complete neutralization. This principle is essential for stoichiometry, which involves calculating the relative quantities of reactants and products in chemical reactions.

Standard Temperature and Pressure (STP)

Standard Temperature and Pressure (STP) is a reference point in chemistry used to easily communicate about gas behaviors and reactions. STP is defined as a temperature of \(273.15\, \text{K}\) (\(0\, \text{°C}\)) and a pressure of \(1\, \text{atm}\). At STP, one mole of an ideal gas occupies \(22.4\, \text{dm}^3\).
In the context of the exercise, this standard allows us to effectively calculate the amount of \(\text{CO}_2\) gas produced during the decomposition of calcium carbonate. Knowing the volume of the gas and understanding that it behaves ideally under these conditions, we used this standard to calculate the number of moles of \(\text{CO}_2\) as \(0.5\). This straightforward correlation between volume and moles simplifies stoichiometric calculations, making STP a valuable concept in both theoretical and practical aspects of chemistry.