Problem 80 - A Complete Resource Book in Chemistry for JEE Main

Question

From the complete decomposition of \(20 \mathrm{~g} \mathrm{CaCO}_{3}\) at STP the volume of \(\mathrm{CO}_{2}\) obtained is (a) \(2.24 \mathrm{~L}\) (b) \(4.48 \mathrm{~L}\) (c) \(44.8 \mathrm{~L}\) (d) \(48.4 \mathrm{~L}\)

Step-by-Step Solution

Answer

The volume of \(\mathrm{CO}_2\) obtained is \(4.48 \text{ L}\) (option b).

Step 1: Write the chemical equation

The decomposition of calcium carbonate can be represented by the chemical equation: \[ \mathrm{CaCO}_3(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO}_2(g) \] This tells us that one mole of \(\mathrm{CaCO}_3\) decomposes to produce one mole of \(\mathrm{CO}_2\).

Step 2: Calculate molar masses

The molar mass of \(\mathrm{CaCO}_3\) is calculated as follows: \[ \text{Ca: } 40.08 \text{ g/mol} + \text{C: } 12.01 \text{ g/mol} + \text{O: } 3 \times 16.00 \text{ g/mol} = 100.09 \text{ g/mol} \]

Step 3: Determine moles of CaCO3

To find the number of moles of \(\mathrm{CaCO}_3\), use the formula: \[ \text{Moles of } \mathrm{CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{20 \text{ g}}{100.09 \text{ g/mol}} \approx 0.1998 \text{ moles} \]

Step 4: Relate moles of CO2 to moles of CaCO3

According to the balanced chemical equation, 1 mole of \(\mathrm{CaCO}_3\) yields 1 mole of \(\mathrm{CO}_2\). Thus, 0.1998 moles of \(\mathrm{CaCO}_3\) will produce 0.1998 moles of \(\mathrm{CO}_2\).

Step 5: Calculate volume of CO2 at STP

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L. Hence, the volume of \(0.1998\) moles of \(\mathrm{CO}_2\) is: \[ 0.1998 \text{ moles} \times 22.4 \text{ L/mole} \approx 4.48 \text{ L} \]

Key Concepts

Calcium Carbonate Decomposition

Calcium carbonate, commonly found in chalk, limestone, and marble, decomposes upon heating in a process known as thermal decomposition. This reaction involves breaking down calcium carbonate (\(\text{CaCO}_3\)) into calcium oxide (\(\text{CaO}\)) and carbon dioxide (\(\text{CO}_2\)). The chemical equation for this decomposition is: \[\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)\]

• **Calcium Carbonate (\(\text{CaCO}_3\)):** A common and naturally occurring substance, often used in construction and manufacturing.
• **Calcium Oxide (\(\text{CaO}\)):** Also known as quicklime, used in cement and as a chemical reagent.
• **Carbon Dioxide (\(\text{CO}_2\)):** A gas found in the Earth's atmosphere, used by plants in photosynthesis.

This reaction is crucial in various industrial processes, like cement production. Understanding it helps students connect theoretical chemistry concepts with practical applications.

Molar Volume of Gases

Gases have specific volume attributes that allow us to predict their behavior under standard conditions, commonly defined as 0°C (273.15 K) and 1 atm pressure. At Standard Temperature and Pressure (STP), one mole of an ideal gas occupies a volume of 22.4 liters. This is known as the molar volume of a gas.The molar volume is particularly useful in calculations involving gases because it allows chemists to predict the volumes of reactants and products involved in chemical reactions. It provides a standardized way to relate moles of gas to volume, which simplifies experiments and predictions in gas chemistry. This concept is critical when calculating the volume of gaseous products, like \(\text{CO}_2\), obtained from reactions such as the decomposition of \(\text{CaCO}_3\).

Mole Concept

The mole is a fundamental unit in chemistry that relates the number of particles in a sample to its mass. One mole of any substance contains Avogadro's number of entities, which is approximately \(6.022 \times 10^{23}\). Using moles, we can convert between mass and number of atoms or molecules. This bridges the gap between microscopic particles and macroscopic masses we handle in the laboratory.To use the mole concept in calculations, such as finding out how many moles of \(\text{CaCO}_3\) are in a 20-gram sample, you divide the mass by the molar mass:
\[\text{Moles of } \text{CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{20 \text{ g}}{100.09 \text{ g/mol}}\]This yields approximately 0.1998 moles. Mastering this concept allows students to decipher the amounts of reactants/products in chemical equations and deeply understand matter and its interactions.

Stoichiometry

Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. By using the balanced chemical equation, stoichiometry helps in calculating how much product can be obtained from a given amount of reactant, or vice versa. In the decomposition of \(\text{CaCO}_3\), stoichiometry indicates that one mole of \(\text{CaCO}_3\) produces one mole of \(\text{CO}_2\). Thus, if 0.1998 moles of \(\text{CaCO}_3\) are decomposed, we know that 0.1998 moles of \(\text{CO}_2\) will be produced due to this molar ratio.

Using the concept of molar volume at STP, we can convert moles of \(\text{CO}_2\) to volume:
\[0.1998 \text{ moles} \times 22.4 \text{ L/mole} \approx 4.48 \text{ L}\]It illustrates the practical application of theoretical principles and strengthens problem-solving skills, critical to mastering chemistry.