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 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.

Step 1: Understand the Reaction

The reaction between sodium carbonate (\( \text{Na}_2\text{CO}_3 \)) and hydrochloric acid (\( \text{HCl} \)) produces carbon dioxide (\( \text{CO}_2 \)). This chemical reaction can be represented as: \[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O} \] In this reaction, one mole of \( \text{Na}_2\text{CO}_3 \) produces one mole of \( \text{CO}_2 \).

Step 2: Calculate Moles of Sodium Carbonate

First, determine the moles of \( \text{Na}_2\text{CO}_3 \) using its mass and molar mass. The moles are calculated using the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \] \[ \text{moles of } \text{Na}_2\text{CO}_3 = \frac{2.12 \, \text{grams}}{106 \, \text{g/mol}} = 0.02 \, \text{mol} \]

Step 3: Relation Between Sodium Carbonate and Carbon Dioxide

From the balanced chemical equation, 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 CO2 at STP

At STP, 1 mole of any gas occupies a volume of 22.4 liters. Hence, the volume of \( \text{CO}_2 \) produced is: \[ \text{Volume of } \text{CO}_2 = 0.02 \, \text{mol} \times 22.4 \, \text{L/mol} = 0.448 \, \text{L} \]

Step 5: Conclusion

Thus, the volume of \( \text{CO}_2 \) liberated at STP is 0.448 liters. Therefore, the correct answer is \((c) \; 0.448\).

Key Concepts

Molar Mass Calculation

Understanding molar mass is crucial for calculating moles from a given mass. Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). To find the molar mass of a compound like sodium carbonate (\(\text{Na}_2\text{CO}_3\)), we need to sum the atomic masses of all atoms present in the compound.

• Sodium (Na) has an atomic mass of approximately 23 g/mol. Since there are two sodium atoms in Na₂CO₃, it contributes \(2 \times 23 = 46 \, \text{g/mol}\) to the molar mass.
• Carbon (C) contributes 12 g/mol because there is one carbon atom.
• Oxygen (O), with three atoms, contributes \(3 \times 16 = 48 \, \text{g/mol}\)

Adding these together, we find that the molar mass of Na₂CO₃ is 106 g/mol. This value allows us to convert grams of a substance to moles using the formula: \[\text{Moles} = \frac{\text{Mass (in g)}}{\text{Molar Mass (in g/mol)}}\] For example, given 2.12 grams of sodium carbonate, the calculation is straightforward with this theorem.

Chemical Reactions

Chemical reactions involve transforming substances into new products. They are represented by balanced equations that ensure the conservation of mass and atoms. In our example, sodium carbonate reacts with hydrochloric acid to produce carbon dioxide, water, and sodium chloride. The balanced reaction is: \[\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O}\] It's important to observe the stoichiometry, which is the mole ratio of reactants and products. Here:- 1 mole of sodium carbonate reacts with 2 moles of hydrochloric acid,- Produces 1 mole of carbon dioxide, 1 mole of water, and 2 moles of sodium chloride.This stoichiometric balance allows us to determine the relation between the amounts of reactants used and products formed, which is key for quantifying reactions.

Gas Laws

Gas laws describe the behavior of gases under different conditions. One crucial aspect is that at Standard Temperature and Pressure (STP), which is 0°C and 1 atm pressure, 1 mole of any ideal gas occupies a volume of 22.4 liters.In our calculation involving the reaction of sodium carbonate with hydrochloric acid, we need to find the volume of carbon dioxide (\text{CO}_2) produced. At STP:- 1 mole of \text{CO}_2 has a volume of 22.4 liters.Given that 0.02 moles of \text{CO}_2 were formed in the reaction, the volume is calculated by:\[0.02 \, \text{mol} \times 22.4 \, \text{L/mol} = 0.448 \, \text{L}\]Understanding gas laws allows for precise calculations when dealing with gaseous products, particularly in conditions defined as STP.

Mole Concept

The mole concept is fundamental in chemistry for measuring substance amounts. One mole corresponds to Avogadro's number, \(6.022 \times 10^{23}\) entities, whether they're atoms, molecules, or ions.This concept links measurable quantities like mass to numbers of atoms and molecules. For instance:- Sodium carbonate's molar mass is 106 g/mol, meaning 106 grams contain one mole.In the original problem:- Given 2.12 grams of sodium carbonate, we calculate moles using the molar mass. -\[\text{Moles of } \text{Na}_2\text{CO}_3 = \frac{2.12 \, \text{g}}{106 \, \text{g/mol}} = 0.02 \, \text{mol}\] Knowing moles helps find quantities involved in chemical reactions as shown by the equation from the problem. It enables linking of macroscopic measurements to molecular-scale interactions.