Problem 111 - A Complete Resource Book in Chemistry for JEE Main

Question

\({ }^{*} \mathrm{X}\) ' gram of calcium carbonate was completely burnt in air. The weight of solid residue formed is \(28 \mathrm{~g}\). What is the value of ' \(\mathrm{X}\) ' (in grams)? (a) 50 (b) 100 (c) 150 (d) 200

Step-by-Step Solution

Answer

The value of '\(X\)' is 50 grams.

Step 1: Write the Chemical Equation

The chemical reaction involved is the decomposition of calcium carbonate (CaCO\(_3\)) into calcium oxide (CaO) and carbon dioxide (CO\(_2\)) upon heating. The balanced equation is: \[\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)\]

Step 2: Determine Molar Masses

Calculate the molar masses of the compounds involved: - Molar mass of CaCO\(_3\) = 40 (Ca) + 12 (C) + 3×16 (O) = 100 g/mol. - Molar mass of CaO = 40 (Ca) + 16 (O) = 56 g/mol.

Step 3: Relate Molar Mass to Reaction

From the balanced equation, 100 g of CaCO\(_3\) produces 56 g of CaO according to their molar masses.

Step 4: Use Given Data and Set Up Ratio

We are given that the solid residue (CaO) weighs 28 g. Set up a proportion based on the molar relationship:\[\frac{100 \text{ g of CaCO}_3}{56 \text{ g of CaO}} = \frac{X \text{ g of CaCO}_3}{28 \text{ g of CaO}}\]

Step 5: Solve for X

Cross-multiply to solve for \(X\):\[X = \frac{100}{56} imes 28 = 50 \text{ g}\]

Step 6: Check Answer Against Options

The calculated \(X\) is 50 g. Verify this with the given options: (a) 50, (b) 100, (c) 150, (d) 200. The correct option is (a) 50.

Key Concepts

Chemical Decomposition

In the world of chemistry, decomposition reactions are a fundamental type of chemical reaction. These involve breaking down a single substance into two or more substances. A well-known example can be observed in the decomposition of calcium carbonate (CaCO\(_3\)). This is the process used in the provided exercise. In a decomposition reaction, the original compound is subjected to conditions such as heat, leading to its breakdown.

During the decomposition of calcium carbonate, it is heated and breaks down into calcium oxide (CaO) and carbon dioxide (CO\(_2\)). This reaction is represented by the balanced chemical equation:

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

This showcases a direct transformation, where the reactant (CaCO\(_3\)) decomposes to form two products. Understanding this concept helps in predicting the products formed when a compound disintegrates under specific conditions.

It is crucial to note that decomposition reactions often require energy input in the form of heat, as seen in this example. This reaction is not only common in chemistry labs but also has significant industrial applications, particularly in the production of lime and cement.

Calcium Carbonate

Calcium carbonate is a chemical compound that can be found in many forms such as limestone, marble, and chalk. In chemistry, it is known for its role in various reactions, particularly decomposition under heat. Calcium carbonate has the chemical formula CaCO\(_3\). It is comprised of one calcium (Ca) atom, one carbon (C) atom, and three oxygen (O) atoms.

The molar mass of calcium carbonate is calculated by summing the atomic masses of its constituent atoms:

• Calcium (Ca) has an atomic mass of approximately 40 g/mol.
• Carbon (C) has an atomic mass of roughly 12 g/mol.
• Oxygen (O) has an atomic mass of 16 g/mol, and there are three oxygen atoms, so this contributes approximately 48 g/mol.

Adding these values gives a molar mass of 100 g/mol for CaCO\(_3\). This is essential for stoichiometric calculations, as it relates the amount of substance to mass.

Moreover, understanding the properties and reactions of calcium carbonate is critical in fields such as geology, manufacturing, and environmental science.

Stoichiometry

Stoichiometry is a branch of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It is based on the conservation of mass and the balanced chemical equations. In the given exercise, stoichiometry is used to determine how much calcium carbonate is needed to produce a specified amount of calcium oxide.

In stoichiometry, ratios from the balanced equation are utilized to relate different substances in a reaction. For the decomposition of calcium carbonate, the equation is:

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

Here, the molar ratio of calcium carbonate to calcium oxide is 1:1, meaning 100 g of CaCO\(_3\) will produce 56 g of CaO, based on their molar masses.

In the exercise, we are given a residue weight and asked to find the initial mass of calcium carbonate. This involves setting up a proportion based on known molar relations and solving for the unknown quantity \(X\). The calculation simplified the setup as:

• \( \frac{100 \text{ g of CaCO}_3}{56 \text{ g of CaO}} = \frac{X \text{ g of CaCO}_3}{28 \text{ g of CaO}} \)
• Solving this gives \(X = 50 \text{ g} \).

This demonstrates how stoichiometry allows chemists to predict the outcomes of reactions quantitatively.