Problem 51 - A Complete Resource Book in Chemistry for JEE Main

Question

In the reaction, \(\mathrm{CH}_{4}+2 \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\), the amount of oxygen needed to completely burn \(4.0 \mathrm{~g}\) of \(\mathrm{CH}_{4}\) is (a) 4 (b) \(8 \mathrm{~g}\) (c) \(16 \mathrm{~g}\) (d) \(32 \mathrm{~g}\)

Step-by-Step Solution

Answer

The correct answer is (c) 16 g.

Step 1: Calculate the moles of CH4

First, we need to find out how many moles of \( \mathrm{CH}_4 \) are in \( 4.0 \ \mathrm{g} \) of \( \mathrm{CH}_4 \). The molar mass of \( \mathrm{CH}_4 \) is approximately \( 16 \ \mathrm{g/mol} \). Therefore, the number of moles is given by:\[ \text{moles of } \mathrm{CH}_4 = \frac{4.0\ \mathrm{g}}{16\ \mathrm{g/mol}} = 0.25 \ \mathrm{mol} \]

Step 2: Relate moles of CH4 to moles of O2

According to the balanced reaction \( \mathrm{CH}_{4} + 2 \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2} + 2 \mathrm{H}_{2}\mathrm{O} \), 1 mole of \( \mathrm{CH}_{4} \) reacts with 2 moles of \( \mathrm{O}_2 \). Therefore, \( 0.25 \ \mathrm{mol} \) of \( \mathrm{CH}_{4} \) will require:\[ 0.25 \times 2 = 0.50 \ \mathrm{mol} \text{ of } \mathrm{O}_2 \]

Step 3: Convert moles of O2 to grams

The molar mass of \( \mathrm{O}_2 \) is approximately \( 32 \ \mathrm{g/mol} \). Thus, \( 0.50 \ \mathrm{mol} \) of \( \mathrm{O}_2 \) is:\[ 0.50 \times 32 \ \mathrm{g/mol} = 16 \ \mathrm{g} \]

Step 4: Select the correct option

From our calculations, we need \( 16 \ \mathrm{g} \) of \( \mathrm{O}_2 \) to completely react with \( 4.0 \ \mathrm{g} \) of \( \mathrm{CH}_4 \). Therefore, the correct option is (c) 16 \ \mathrm{g}.

Key Concepts

Molar Mass Calculation

Molar mass is a fundamental concept in chemistry that connects the mass of a substance to the amount of matter it contains. It tells us how many grams are in one mole of a substance, which is important for converting between mass and moles. To calculate the molar mass of a compound, you need to sum up the atomic masses of each element that makes up the compound. For instance, methane (\( \mathrm{CH}_4 \)) consists of one carbon atom (with an atomic mass of about \( 12\ \mathrm{g/mol} \)) and four hydrogen atoms (each about \( 1\ \mathrm{g/mol} \)).
Add these together:

• Carbon: \( 12\\mathrm{g/mol} \)
• Hydrogen: \( 1\mathrm{g/mol} \times 4 = 4\mathrm{g/mol} \)

Total molar mass of \( \mathrm{CH}_4 \) is \( 12 + 4 = 16 \mathrm{g/mol} \).
This tells us that one mole of methane weighs \( 16 \mathrm{g} \). Understanding this calculation helps us move easily between using the mass of a substance and the amounts in moles, crucial for reacting these substances in chemical equations.

Mole Concept

The mole is a core concept in chemistry that allows us to quantify the number of particles in a substance easily and effectively. One mole equals \(6.022\times10^{23}\) of anything, whether atoms, molecules, or ions, a number called Avogadro's number. This large number allows chemists to work with substances at the macroscopic level in the lab rather than individual atoms or molecules.
When given a mass and you want to find out how many moles you have, divide the mass by the molar mass. For example, with 4.0 grams of \( \mathrm{CH}_4 \):

• Molar mass of \( \mathrm{CH}_4 \) is \( 16 \mathrm{g/mol} \)
• Number of moles: \( \frac{4.0g}{16 \mathrm{g/mol}} = 0.25 \mathrm{mol} \)

Understanding the mole concept is essential for accurately using amounts of substances in chemical reactions and calculations, ensuring precise reactions and outcomes in chemical processes.

Balanced Chemical Equations

A chemical equation represents a chemical reaction where reactants are transformed into products. A balanced chemical equation has equal numbers of each type of atom on both sides of the equation, reflecting the conservation of mass. For instance, the balanced equation \( \mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2 \mathrm{O} \) shows that one molecule of methane reacts with two molecules of oxygen, yielding one molecule of carbon dioxide and two molecules of water.
Balancing is critical because it ensures that the calculation of reactants and products is accurate. It also helps in determining the stoichiometric relationships, meaning the exact proportions of substances required and produced. For the given reaction, by understanding the balanced equation, you know:

• 1 mole of \( \mathrm{CH}_4 \) reacts with 2 moles of \( \mathrm{O}_2 \)
• Using this ratio, 0.25 moles of \( \mathrm{CH}_4 \) would require 0.5 moles of \( \mathrm{O}_2 \)

Balancing equations ensures that you have the right amounts of reactants for the complete reaction and is crucial for successfully performing chemical calculations and experiments.