Problem 29 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

\(\mathrm{NaClO}_{3}\) is used, even in spacecrafts, to produce \(\mathrm{O}_{2}\). The daily consumption of pure \(\mathrm{O}_{2}\) by a person is \(492 \mathrm{~L}\) at 1 atm, \(300 \mathrm{~K}\). How much amount of \(\mathrm{NaClO}_{3}\), in grams, is required to produce \(\mathrm{O}_{2}\) for the daily consumption of a person at \(1 \mathrm{~atm}, 300 \mathrm{~K}\) ? \(\mathrm{NaClO}_{3}(\mathrm{~s})+\mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{O}_{2}(\mathrm{~g})+\mathrm{NaCl}(\mathrm{s})+\mathrm{FeO}(\mathrm{s}) \mathrm{R}=0 . \overline{082 \mathrm{Latm}} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\) [Main Jan. 08, 2020 (II)]

Step-by-Step Solution

Answer

2130 grams of \( \text{NaClO}_3 \) is needed.

Step 1: Understand the reaction

The given reaction is \( \text{NaClO}_3(s) + \text{Fe}(s) \rightarrow \text{O}_2(g) + \text{NaCl}(s) + \text{FeO}(s) \). This tells us that 1 mole of \( \text{NaClO}_3 \) produces a certain amount of \( \text{O}_2 \).

Step 2: Use the Ideal Gas Law

Use the ideal gas law to find the number of moles of \( \text{O}_2 \) needed. The formula is \( PV = nRT \). Here, \( P = 1 \text{ atm} \), \( V = 492 \text{ L} \), \( R = 0.0821 \text{ L atm K}^{-1} \text{mol}^{-1} \), and \( T = 300 \text{ K} \). Solve for \( n \), the number of moles of \( \text{O}_2 \): $$ n = \frac{PV}{RT} = \frac{1 \times 492}{0.0821 \times 300} \approx 20.0 \text{ moles} $$

Step 3: Determine moles of NaClO3 required

From the balanced chemical equation, every 1 mole of \( \text{NaClO}_3 \) produces 1 mole of \( \text{O}_2 \). Therefore, 20 moles of \( \text{O}_2 \) require 20 moles of \( \text{NaClO}_3 \).

Step 4: Calculate mass of NaClO3

The molar mass of \( \text{NaClO}_3 \) is approximately \( 23+35.5+3\times16 = 106.5 \text{ g/mol} \). So, the mass needed is: $$ \text{mass} = 20 \text{ moles} \times 106.5 \text{ g/mol} = 2130 \text{ grams} $$

Key Concepts

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In the given exercise, the reaction is: \(\mathrm{NaClO}_3(\mathrm{s}) + \mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{O}_2(\mathrm{g}) + \mathrm{NaCl}(\mathrm{s}) + \mathrm{FeO}(\mathrm{s})\). Here, sodium chlorate \(\mathrm{NaClO}_3\) is used to generate oxygen \(\mathrm{O}_2\).
This reaction is essential for understanding how substances like sodium chlorate can play a role in generating vital gases such as oxygen, especially in controlled environments like spacecraft.
The decomposition of \(\mathrm{NaClO}_3\) when reacting with iron results in the production of oxygen gas, table salt \((\mathrm{NaCl})\), and iron(II) oxide \((\mathrm{FeO})\). This reaction is balanced, meaning the number of atoms for each element is equal on both sides of the equation. This principle ensures that mass is conserved during the reaction.

Mole Concept

The mole concept is crucial in chemistry, serving as a bridge between the atomic world and the practical scale we observe. One mole is defined as \(6.022 \times 10^{23}\) atoms, molecules, or ions of a substance, and it provides a way to count particles by weighing them.
In the context of the exercise, the ideal gas law formula \( PV = nRT \) is applied to find the number of moles of oxygen gas. Using the given conditions:

• Pressure \( P = 1 \text{ atm} \)
• Volume \( V = 492 \text{ L} \)
• Temperature \( T = 300 \text{ K} \)
• Ideal Gas Constant \( R = 0.0821 \text{ L atm K}^{-1} \text{mol}^{-1} \)

By solving \( n = \frac{PV}{RT} \), we determine there are approximately 20 moles of \(\mathrm{O}_2\) needed. This links the macroscopic volume of gas to the microscopic amount of substance.

Stoichiometry

Stoichiometry connects the quantitative aspects of chemical formulas and reactions. It enables the calculation of reactants and products in chemical reactions.
In the balanced chemical equation, it's evident that each mole of \(\mathrm{NaClO}_3\) yields one mole of \(\mathrm{O}_2\). Since 20 moles of \(\mathrm{O}_2\) are required, 20 moles of \(\mathrm{NaClO}_3\) will be necessary. Understanding mole-to-mole relationships is pivotal for stoichiometric calculations.
The next step is to convert moles of \(\mathrm{NaClO}_3\) to grams. The molar mass of \(\mathrm{NaClO}_3\) is calculated as follows:

• Sodium: 23 g/mol
• Chlorine: 35.5 g/mol
• Oxygen: \(3 \times 16 = 48 \) g/mol

Total molar mass \(= 106.5 \text{ g/mol}\).
Thus, the mass of \(\mathrm{NaClO}_3\) needed is calculated by multiplying the number of moles by the molar mass, giving \(2130 \text{ grams}\). This step is essential for any chemist looking to predict the quantities of reactants required or products formed in a chemical reaction.