Problem 152 - A Complete Resource Book in Chemistry for JEE Main

Question

Oleum is considered as a solution of \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\), which is obtained by passing \(\mathrm{SO}_{3}\) in concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\). When \(100 \mathrm{~g}\) sample of oleum is diluted with desired weight of \(\mathrm{H}_{2} \mathrm{O}\), then the total mass of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) obtained after dilution is known as \% labeling in oleum. For example, an oleum labeled as '109\% \(\mathrm{H}_{2} \mathrm{SO}_{4}{\underline{\phantom{xx}}}^{\prime}\) means the \(109 \mathrm{~g}\) total mass of pure \(\mathrm{H}_{2} \mathrm{SO}_{4}\) will be formed when \(100 \mathrm{~g}\) of oleum is diluted by \(9 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) which combines with all the free \(\mathrm{SO}_{3}\) present in oleum to form \(\mathrm{H}_{2} \mathrm{SO}_{4} \mathrm{as} \mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}\). \(1 \mathrm{~g}\) of oleum sample is diluted with water. The solution required \(54 \mathrm{~mL}\) of \(0.4 \mathrm{~N} \mathrm{NaOH}\) for complete neutralization. The \% of free \(\mathrm{SO}_{3}\) in the sample is (a) 20 (b) 52 (c) 26 (d) none of these

Step-by-Step Solution

Answer

The percentage of free \( \mathrm{SO}_{3} \) is not among the choices; it's 86.4\%.

Step 1: Understand the Reaction

When oleum is diluted with water, free SO₃ combines with water as \( \mathrm{SO}_{3} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4} \). This indicates that the amount of free SO₃ can be determined via neutralization with NaOH.

Step 2: Use the Neutralization Equation

The neutralization reaction can be written as \( \mathrm{H}_{2} \mathrm{SO}_{4} + 2\mathrm{NaOH} \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{4} + 2\mathrm{H}_{2} \mathrm{O} \). Since 1 mole of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) reacts with 2 moles of \( \mathrm{NaOH} \), the normality of \( \mathrm{NaOH} \) doubles the moles of \( \mathrm{NaOH} \) reacting.

Step 3: Calculate Moles of NaOH Used

The volume of \( \mathrm{NaOH} \) is 54 mL or 0.054 L. With a normality of 0.4 N, the moles of NaOH reacting is \( 0.054 \times 0.4 = 0.0216 \) equivalents.

Step 4: Convert NaOH to Moles of SO₃

Given that \( 1 \mathrm{mol} \) of \( \mathrm{SO}_{3} \) neutralizes 2 equivalents of \( \mathrm{NaOH} \), the moles of SO₃ in 1 g of oleum is \( \frac{0.0216}{2} = 0.0108 \) mol. This implies free SO₃ contributes directly to the moles of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) after dilution.

Step 5: Calculate Mass of Free SO₃

The molar mass of \( \mathrm{SO}_{3} \) is 80 g/mol. Thus, the mass of SO₃ is \( 0.0108 \times 80 = 0.864 \mathrm{~g} \).

Step 6: Find Percentage by Mass of Free SO₃

Since 1 g of the oleum sample contains 0.864 g of free SO₃, the percentage of free \( \mathrm{SO}_{3} \) in the sample is \( \left( \frac{0.864}{1.0} \right) \times 100 \% = 86.4 \% \).

Key Concepts

Neutralization Reaction

A neutralization reaction occurs when an acid and a base react to form water and a salt. This is a very common reaction in chemistry and provides the backbone for various calculations and chemical engineering processes. In our context, sulfuric acid (\(\mathrm{H}_{2}\mathrm{SO}_{4}\)) is formed from the reaction of sulfur trioxide (\(\mathrm{SO}_{3}\)) with water.
For this phenomenon, the general equation can be represented as follows:

• \(\mathrm{SO}_{3} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}\)

To analyze the amount of free \(\mathrm{SO}_{3}\) present, a neutralization reaction with sodium hydroxide (\(\mathrm{NaOH}\)) can be performed. The equation for the neutralization of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) with \(\mathrm{NaOH}\) is depicted as:

• \(\mathrm{H}_{2}\mathrm{SO}_{4} + 2\mathrm{NaOH} \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{4} + 2 \mathrm{H}_{2}\mathrm{O}\)

Understanding these reactions allows us to calculate how much of the substance reacts under given conditions, an essential in chemical analysis.

Sulfur Trioxide

Sulfur trioxide (\(\mathrm{SO}_{3}\)) is a compound critical in the manufacturing of sulfuric acid. It's a key ingredient in the production of oleum, an important intermediate in the industry.
In its gaseous state, sulfur trioxide is highly reactive and when mixed with water, it rapidly forms sulfuric acid. This reaction is fundamental in creating the world’s most consumed industrial chemical. Despite its industrial importance, \(\mathrm{SO}_{3}\) is highly corrosive and needs to be handled with care.
In practice, \(\mathrm{SO}_{3}\) can be efficiently converted into sulfuric acid using controlled conditions. The handling of this substance, especially during the oleum process, requires precision to ensure safety and effectiveness in production.

Sulfuric Acid

Sulfuric acid (\(\mathrm{H}_{2}\mathrm{SO}_{4}\)) is one of the most important substances in chemical manufacturing. It's known for its highly acidic and hygroscopic nature, meaning it can absorb water from the air. In the industrial production process, oleum, which contains dissolved \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2}\mathrm{SO}_{4}\), is a crucial step in enhancing the concentration of sulfuric acid.
This concentrated sulfuric acid is used extensively: in fertilizers, chemical synthesis, oil refining, and in wastewater processing, to name a few applications.
The labeling of oleum with percentages signifies the potential yield of pure \(\mathrm{H}_{2}\mathrm{SO}_{4}\) that can be achieved through full reaction and dilution with water.

Molar Mass Calculations

Understanding molar mass is essential in chemistry as it allows you to convert between the mass of a substance and the amount in moles. The molar mass is the mass of one mole of a given substance.
For instance, the molar mass of sulfur trioxide (\(\mathrm{SO}_{3}\)) is 80 g/mol, calculated by adding the atomic masses of sulfur and oxygen (32 for sulfur and 16 per oxygen atom).
To determine the amount of a substance required in a reaction, molar mass calculations transform weight measurements into meaningful data.
In our exercise, knowing the molar mass helps calculate how much \(\mathrm{SO}_{3}\) is present from the experimental data, thus finding the percentage composition by mass.