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

The number of moles of sodium bicarbonate formed is 0.2 moles.

Step 1: Calculate the moles of initial compound

First, we need to calculate the number of moles of \(\text{CaCO}_3\) using the given mass and molar mass.Given:- Mass of \(\text{CaCO}_3 = 10\, \text{g}\)- Molar mass of \(\text{CaCO}_3 = 100\, \text{g/mol}\)Number of moles of \(\text{CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{10}{100} = 0.1\, \text{mol}\).

Step 2: Decompose the compound

Upon decomposition, \(\text{CaCO}_3\) breaks down according to the reaction:\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]Here, \(\text{X}\) is \(\text{CO}_2\). For every mole of \(\text{CaCO}_3\) decomposed, 1 mole of \(\text{CO}_2\) and 1 mole of \(\text{CaO}\) are produced. Thus, \(0.1\, \text{mol}\) of \(\text{CO}_2\) is formed.

Step 3: React \(\text{X}\) with sodium carbonate

The \(\text{CO}_2\) produced reacts with the sodium carbonate solution:\[ \text{CO}_2 + \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \rightarrow 2 \text{NaHCO}_3 \]According to the reaction stoichiometry, 1 mole of \(\text{CO}_2\) reacts with 1 mole of \(\text{Na}_2\text{CO}_3\) to form 2 moles of \(\text{NaHCO}_3\).

Step 4: Calculate moles of sodium bicarbonate formed

We calculated in Step 2 that \(0.1\, \text{mol}\) of \(\text{CO}_2\) is formed. From the stoichiometric equation above, \(0.1\, \text{mol}\) of \(\text{CO}_2\) will react with \(0.1\, \text{mol}\) of \(\text{Na}_2\text{CO}_3\) to produce \(2 \times 0.1 = 0.2\, \text{mol}\) of \(\text{NaHCO}_3\).

Key Concepts

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. Understanding stoichiometry helps us predict how much of a product will form or how much of a reactant is required in a given reaction.

In the context of the problem, stoichiometry is used to calculate the moles of each compound involved in the decomposition and subsequent reactions. This is done by using the balanced chemical equation and the established relationships between reactants and products.

For example:
- Calculating the number of moles involves using the formula: \( ext{Number of moles} = rac{ ext{mass}}{ ext{molar mass}} \).
- The balanced chemical equation then tells us that 1 mole of \( ext{CaCO}_3\) decomposes to form 1 mole of \( ext{CO}_2\). Using stoichiometry, you can determine reaction quantities and the resulting chemical products.

Calcium Carbonate Decomposition

Calcium carbonate (CaCO_3) decomposition is a specific type of chemical reaction known as thermal decomposition. In this process, a single compound breaks down into simpler substances when heated, often releasing a gas.

In the given exercise, when \( ext{CaCO}_3\) is heated, it decomposes to form calcium oxide (CaO) and carbon dioxide (CO_2):

\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]

This equation tells us that for every mole of \( ext{CaCO}_3\) decomposed, 1 mole of \( ext{CaO}\) and 1 mole of \( ext{CO}_2\) are formed. This type of reaction is an important process in industries, such as lime production, where calcium oxide is a crucial product.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry refers to the relationship between the amounts of reactants and products in a chemical reaction. The balanced chemical equation is key to understanding these relationships.

In the exercised problem, stoichiometry plays a role in calculating the amount of sodium bicarbonate (NaHCO_3) that is formed after \( ext{CO}_2\), a product of \( ext{CaCO}_3\) decomposition, reacts with sodium carbonate (Na_2CO_3).

The reaction is:

\[ \text{CO}_2 + \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \rightarrow 2 \text{NaHCO}_3 \]

This balanced equation indicates that 1 mole of \( ext{CO}_2\) reacts with 1 mole of \( ext{Na}_2CO_3\) to produce 2 moles of \( ext{NaHCO}_3\). Thus, if \(0.1\) moles of \( ext{CO}_2\) are available, it will react to form \(0.2\) moles of \( ext{NaHCO}_3\), demonstrating the principles of stoichiometry.