Problem 107 - A Complete Resource Book in Chemistry for JEE Main

Question

\(10 \mathrm{~g}\) of \(\mathrm{CaCO}_{3}\) is completely decomposed to \(\mathrm{X}\) and \(\mathrm{CaO} . \mathrm{X}\) is passed into an aqueous solution containing one mole of sodium carbonate. What is the number of moles of sodium bicarbonate formed? (Mol. wt of \(\mathrm{CaCO}_{3}=100, \mathrm{Na}_{2} \mathrm{CO}_{3}=106\) \(\left.\mathrm{NaHCO}_{3}=84\right)\) (a) \(0.010\) (b) \(0.2\) (c) \(0.4\) (d) 10

Step-by-Step Solution

Answer

0.2 moles of sodium bicarbonate are formed (option b).

Step 1: Write the Decomposition Reaction

When calcium carbonate (\( \text{CaCO}_3 \)) decomposes, it forms calcium oxide (\( \text{CaO} \)) and carbon dioxide gas (\( \text{CO}_2 \)). The equation for the reaction is: \[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]

Step 2: Calculate the Moles of Calcium Carbonate Decomposed

Given that the molecular weight of \( \text{CaCO}_3 \) is 100 g/mol, the number of moles of 10 g of \( \text{CaCO}_3 \) is calculated by: \[ \text{Moles of CaCO}_3 = \frac{10 \text{ g}}{100 \text{ g/mol}} = 0.1 \text{ mol} \]

Step 3: Determine Moles of Carbon Dioxide Produced

From the decomposition reaction, 1 mole of \( \text{CaCO}_3 \) produces 1 mole of \( \text{CO}_2 \). Therefore, 0.1 mol of \( \text{CaCO}_3 \) will produce 0.1 mol of \( \text{CO}_2 \).

Step 4: Write the Reaction with Sodium Carbonate

The \( \text{CO}_2 \) reacts with sodium carbonate (\( \text{Na}_2\text{CO}_3 \)) to form sodium bicarbonate (\( \text{NaHCO}_3 \)). The reaction is: \[ \text{CO}_2 + \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \rightarrow 2 \text{NaHCO}_3 \]

Step 5: Calculate Moles of Sodium Bicarbonate Formed

Since 1 mole of \( \text{CO}_2 \) reacts with 1 mole of \( \text{Na}_2\text{CO}_3 \) to form 2 moles of \( \text{NaHCO}_3 \), 0.1 moles of \( \text{CO}_2 \) will react with sodium carbonate to form \( 2 \times 0.1 = 0.2 \) moles of \( \text{NaHCO}_3 \).

Key Concepts

Decomposition Reaction

In chemistry, a decomposition reaction is a type of chemical reaction where a single compound breaks down into two or more simpler substances.
This process involves the breaking of chemical bonds. Decomposition reactions often require an input of energy in the form of heat, light, or electricity.
For example, when heating calcium carbonate (CaCO₃), it decomposes to form calcium oxide (CaO) and carbon dioxide (CO₂).
The chemical equation for this decomposition reaction is:

• CaCO₃ → CaO + CO₂

Here, you can see how one reactant breaks down into two products. This fundamental concept is crucial in many industrial processes and laboratory settings.

Moles Calculation

Understanding moles is key to grasping stoichiometry in chemistry. A mole is a unit that measures the amount of a substance, based on the number of atoms or molecules present, similar to how a dozen measures 12 items.
To calculate moles, you use the formula: \[ \text{Moles} = \frac{\text{Mass of the substance (g)}}{\text{Molecular weight (g/mol)}} \] In this exercise, to find the moles of calcium carbonate (CaCO₃), you divide its mass by its molecular weight. Given that 10 g of CaCO₃ is used, with a molecular weight of 100 g/mol:

• Moles of CaCO₃ = \( \frac{10 \text{ g}}{100 \text{ g/mol}} = 0.1 \text{ mol} \)

This calculation enables you to proceed with the stoichiometry of the given reactions, predicting how much product is formed.

Chemical Reaction Equations

Chemical reaction equations represent the starting materials and resulting products of a reaction, showing the transformation that occurs. These equations must be balanced to ensure the law of conservation of mass is upheld; meaning, the number of atoms for each element is the same on both sides.
In our scenario, when calcium carbonate decomposes:

• CaCO₃ → CaO + CO₂

Each component of the equation is essential: on the left, you have the reactant (CaCO₃), and on the right, the products (CaO and CO₂). Furthermore, when CO₂ reacts with sodium carbonate (Na₂CO₃) in water, it forms sodium bicarbonate (NaHCO₃):

• CO₂ + Na₂CO₃ + H₂O → 2 NaHCO₃

Here, stoichiometry ensures that the quantities of all substances involved are correctly represented, leading to accurate predictions about product formation.

Sodium Bicarbonate Formation

Sodium bicarbonate, also known as baking soda, forms when carbon dioxide interacts with sodium carbonate in an aqueous solution.
The reaction can be expressed by the equation:

• CO₂ + Na₂CO₃ + H₂O → 2 NaHCO₃

In this context, one mole of CO₂ reacts with one mole of sodium carbonate to produce two moles of sodium bicarbonate. This stoichiometric relationship involves calculating the products based on the reactant quantities.
For example, with 0.1 moles of CO₂ from the decomposition of calcium carbonate, reacting with sodium carbonate yields:

• 2 × 0.1 = 0.2 moles of NaHCO₃

This method of calculation is crucial in chemical manufacturing and other applications where precise amounts of a product are required.