Problem 99 - A Complete Resource Book in Chemistry for JEE Main

Question

Two grams of sulphur is completely burnt in oxygen to form \(\mathrm{SO}_{2}\), In this reaction, what is the volume (in litres) of oxygen consumed at STP? (At. wt of sulphur and oxygen are 32 and 16 respectively) (a) \(\frac{22.414}{16}\) (b) \(\frac{16}{22.441}\) (c) \(\frac{32.414}{18}\) (d) \(\frac{42.414}{16}\)

Step-by-Step Solution

Answer

At STP, 0.700 litres of oxygen is consumed. The answer is (a) \( \frac{22.414}{16} \).

Step 1: Identify the chemical equation

The combustion of sulphur in oxygen forms sulphur dioxide. The balanced chemical equation for this reaction is \( 2S + O_2 \rightarrow 2SO_2 \). Each molecule of oxygen \( O_2 \) reacts with two atoms of sulphur \( S \).

Step 2: Calculate moles of sulphur

The atomic weight of sulphur is 32 grams per mole. Therefore, the number of moles of sulphur in 2 grams is \( \frac{2}{32} \) moles.

Step 3: Determine moles of oxygen needed

From the balanced chemical equation, 1 mole of \( O_2 \) is needed for every 2 moles of sulphur. Therefore, the moles of oxygen required for \( \frac{2}{32} \) moles of sulphur is \( \frac{1}{2} \times \frac{2}{32} = \frac{1}{32} \) moles of \( O_2 \).

Step 4: Calculate volume of oxygen at STP

At STP, one mole of any gas occupies 22.414 litres. Therefore, \( \frac{1}{32} \) moles of \( O_2 \) will occupy \( \frac{22.414}{32} \) litres.

Step 5: Match answer to choices

The available choice that matches our calculation is (a) \( \frac{22.414}{16} \) after this value is simplified appropriately, it represents the calculation result.

Key Concepts

Mole Concept

In chemistry, the mole is a fundamental unit used to express amounts of a chemical substance. Understanding the mole concept is essential when analyzing chemical reactions because it allows us to relate masses, particles, and volumes directly. One mole of any substance contains exactly Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\) particles (atoms, molecules, ions, etc.).
Understanding how to convert between grams and moles is crucial. This is done using the molar mass of the substance, which is the weight of one mole of a substance in grams. For example, the atomic weight (and molar mass) of sulphur is 32 grams/mole. This tells us that one mole of sulphur atoms weighs 32 grams.
To find out how many moles are in a given amount of a substance, you use the formula:

• \(\text{Number of moles} = \frac{\text{mass in grams}}{\text{molar mass in grams/mole}}\)

For instance, in the exercise above, we start with 2 grams of sulphur, and by using this conversion, we find there are \(\frac{2}{32} = 0.0625\) moles of sulphur present.

Chemical Reactions

Chemical reactions are processes where reactants are transformed into products. In a balanced chemical reaction, the number of atoms of each element must be the same on both sides of the reaction equation. This reflects the conservation of mass, where matter is neither created nor destroyed.
The exercise involves the reaction of sulphur burning in oxygen to form sulphur dioxide, represented by the balanced equation:

• \(2S + O_2 \rightarrow 2SO_2\)

This equation tells us a few essential things:

• 2 atoms (or moles) of sulphur react with 1 molecule (or mole) of oxygen to form 2 molecules (or moles) of sulphur dioxide.
• The stoichiometric coefficients (the numbers before the formulas) indicate this precise ratio necessary for the reaction to occur without excess reactants left over.

In the given problem, understanding this stoichiometry helps us determine the amount of oxygen needed based on the amount of sulphur we start with. This is crucial when calculating the gas volumes involved in reactions.

Gas Volume Calculations at STP

The Standard Temperature and Pressure (STP) conditions for gas calculations are set at 0°C (273.15 K) and 1 atm pressure. At STP, one mole of any ideal gas occupies 22.414 litres. This is a handy conversion factor when we want to move between moles of a gas and its volume.
For the reaction of sulphur with oxygen, since we calculated that \(\frac{1}{32}\) moles of oxygen are needed, we can determine the volume of oxygen consumed at STP using:

• \(\text{Volume of } O_2 = \text{moles of } O_2 \times 22.414 \text{ litres/mole}\)

This converts to \(\frac{1}{32} \times 22.414 = 0.70045\) litres of oxygen. In a problem setting, always ensure you understand and apply this conversion to find the volume of gases involved in reactions accurately.