Problem 135 - A Complete Resource Book in Chemistry for JEE Main

Question

What is the volume (in litres) of \(\mathrm{CO}_{2}\) liberated at STP, when \(2.12\) gram of sodium carbonate (mol. \(\mathrm{wt}=106\) ) is treated with excess dilute \(\mathrm{HCl}\) ? (a) \(11.2\) (b) \(2.12\) (c) \(0.448\) (d) \(4.26\)

Step-by-Step Solution

Answer

The volume of \(\text{CO}_2\) liberated is \(0.448\) liters (option c).

Step 1: Write the Balanced Chemical Equation

When sodium carbonate (\(\text{Na}_2\text{CO}_3\)) reacts with hydrochloric acid (\(\text{HCl}\)), carbon dioxide (\(\text{CO}_2\)) gas is produced. The balanced chemical equation is: \[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \] This tells us that 1 mole of \(\text{Na}_2\text{CO}_3\) produces 1 mole of \(\text{CO}_2\).

Step 2: Calculate Moles of Sodium Carbonate

Given the mass of sodium carbonate is 2.12 grams and its molar mass is 106 g/mol, we can calculate the number of moles: \[ \text{Moles of } \text{Na}_2\text{CO}_3 = \frac{2.12}{106} = 0.02 \text{ moles} \]

Step 3: Determine Moles of Produced Carbon Dioxide

From the reaction stoichiometry, 1 mole of \(\text{Na}_2\text{CO}_3\) produces 1 mole of \(\text{CO}_2\). Thus, 0.02 moles of \(\text{Na}_2\text{CO}_3\) will produce 0.02 moles of \(\text{CO}_2\).

Step 4: Calculate Volume of Carbon Dioxide at STP

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Thus, the volume of \(0.02\) moles of \(\text{CO}_2\) is: \[ \text{Volume of } \text{CO}_2 = 0.02 \times 22.4 = 0.448 \text{ liters} \]

Step 5: Identify the Correct Answer

Compare the calculated volume of \(0.448\) liters with the given options. The closest option is (c) \(0.448\).

Key Concepts

Balanced Chemical Equation

To understand chemical reactions, it is essential to know what a balanced chemical equation is. A balanced equation shows the relationship between the reactants and the products in a chemical reaction. When sodium carbonate \((\text{Na}_2\text{CO}_3)\) reacts with hydrochloric acid \((\text{HCl})\), it produces sodium chloride \((\text{NaCl})\), water \((\text{H}_2\text{O})\), and carbon dioxide \((\text{CO}_2})\).

The reaction can be represented as: \[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]This equation tells us that one mole of sodium carbonate reacts with two moles of hydrochloric acid to produce one mole each of water and carbon dioxide, along with two moles of sodium chloride. The equation is balanced when the number of atoms on the reactant side equals the number on the product side. This balance follows the law of conservation of mass, ensuring that matter is neither created nor destroyed during the reaction.

Moles Calculation

Understanding moles is crucial in chemistry because it allows for the quantification of chemical substances. The mole is a unit that measures the amount of substance based on a given number of atoms, which is Avogadro's number \((6.022 \times 10^{23})\).

To find out how many moles are in a certain mass of a compound, like sodium carbonate, you need its molar mass (the mass of one mole of that compound). For sodium carbonate, the molar mass is 106 g/mol. Given a sample of sodium carbonate weighing 2.12 grams, you can calculate the moles using the formula:

• \[ \text{Moles of } \text{Na}_2\text{CO}_3 = \frac{2.12}{106} = 0.02 \text{ moles} \]

This calculation shows you have 0.02 moles of sodium carbonate in the given mass, which is critically important for stoichiometric calculations in a balanced chemical equation.

Standard Temperature and Pressure (STP)

Standard Temperature and Pressure (STP) are important terms in chemistry. STP refers to a standard set of conditions where:

• Temperature is \(0^\circ\text{C}\) or 273.15 K
• Pressure is 1 atm (atmosphere)

These conditions are commonly used to define the behavior of gases. At STP, one mole of any gas occupies a volume of 22.4 liters. This uniformity allows us to predict and calculate various properties of gases, such as their volume when under standard conditions. Knowing about STP is crucial when dealing with gas reactions, as it provides a basis to compare behaviors and properties across different gases.

Gas Volume Calculation

Calculating the volume of gases like carbon dioxide at STP is an application of the mole concept regarding gases. As given in our exercise, we calculated that 0.02 moles of \(\text{CO}_2\) are produced from the reaction.

At STP, each mole of a gas occupies 22.4 liters. Thus, to find the volume of \(\text{CO}_2\) produced, you multiply the number of moles by 22.4 liters per mole:

• \[ \text{Volume of } \text{CO}_2 = 0.02 \text{ moles} \times 22.4 \text{ liters/mole} = 0.448 \text{ liters} \]

This calculation demonstrates how stoichiometry and the conditions of STP help in determining the volume of gas produced in a chemical reaction. This is extremely helpful in predicting outcomes and planning experiments where gases are involved.