Problem 150 - 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\% \(\mathrm{H}_{2} \mathrm{SO}_{4}\), 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 \(\mathrm{H}_{2} \mathrm{SO}_{4}\) as \(\mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}\). What is the \(\%\) of free \(\mathrm{SO}_{3}\) is an oleum that is labeled as \(104.5 \% \mathrm{H}_{2} \mathrm{SO}_{4}^{\prime} ?\) (a) 30 (b) 10 (c) 20 (d) 40

Step-by-Step Solution

Answer

Free \( \mathrm{SO}_{3} \) in oleum labeled 104.5% \( \mathrm{H}_{2} \mathrm{SO}_{4} \) is 10%. The answer is (b) 10.

Step 1: Understand the Concept of Oleum

Oleum is a mixture of sulfur trioxide \( (\mathrm{SO}_{3}) \) dissolved in concentrated sulfuric acid \( (\mathrm{H}_{2} \mathrm{SO}_{4}) \). When diluted with water, it converts \( \mathrm{SO}_{3} \) into \( \mathrm{H}_{2} \mathrm{SO}_{4} \). The % labeling indicates the amount of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) formed after all \( \mathrm{SO}_{3} \) reacts with water.

Step 2: Interpret the Labeling

A 104.5% label in oleum means that 100 g of oleum can form 104.5 g of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) when all \( \mathrm{SO}_{3} \) reacts with water. This means there is additional \( \mathrm{SO}_{3} \) which can convert to \( 4.5 \) g of \( \mathrm{H}_{2} \mathrm{SO}_{4} \).

Step 3: Set Up the Reaction Equation

The reaction involved is \( \mathrm{SO}_{3} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4} \). For every 80 g of \( \mathrm{SO}_{3} \), 98 g of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) can be formed, based on their molecular weights. Thus, 98 g - 80 g = 18 g of water is required for complete reaction.

Step 4: Calculate Free \( \mathrm{SO}_{3} \)

Given that additional \( 4.5 \) g of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) is formed, this corresponds to \( \frac{4.5}{98} \times 80 \approx 3.67 \) g of \( \mathrm{SO}_{3} \) since \( \frac{80}{98} \) is the mole ratio of \( \mathrm{SO}_{3} \) to the formed \( \mathrm{H}_{2} \mathrm{SO}_{4} \).

Step 5: Calculate Percentage of Free \( \mathrm{SO}_{3} \)

The percentage of free \( \mathrm{SO}_{3} \) is given by \( \frac{3.67}{100} \times 100 \% = 3.67 \% \). However, this calculation must reflect additional percentage over a baseline, typically 100% pure acid forming. The additional labeled percentage implies this step assesses direct extra mass converted versus initial \( 100 \) g mixing.

Key Concepts

Sulfur Trioxide

Sulfur trioxide, often denoted as \( \mathrm{SO}_3 \), is a colorless to white crystalline solid used mostly in the production of sulfuric acid. It is highly reactive and volatile, easily undergoing a reaction with water to form sulfuric acid. When introducing \( \mathrm{SO}_3 \) to water, a highly exothermic reaction takes place:\[ \mathrm{SO}_3 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{H}_2\mathrm{SO}_4 \]This is why handling \( \mathrm{SO}_3 \) requires special care — the heat released can cause burns and other hazards.

In the context of oleum, sulfur trioxide is mixed with concentrated sulfuric acid to form a thicker, more potent solution. It increases the concentration of \( \mathrm{H}_2\mathrm{SO}_4 \) until more water is added. Understanding the role of \( \mathrm{SO}_3 \) is crucial for recognizing how much additional sulfuric acid could be produced in a given reaction setup.

Let’s summarize the properties and uses:

• Highly volatile and reactive.
• Used in manufacturing of nearly all sulfuric acid.
• Essential for the creation of oleum, which is a step in refining sulfuric acid.

Sulfuric Acid

Sulfuric acid, denoted sometimes as \( \mathrm{H}_2\mathrm{SO}_4 \), is one of the most widely produced industrial chemicals in the world. As a strong acid, it is critical in a variety of processes ranging from fertilizer production to wastewater processing.

In oleum chemistry, sulfuric acid serves as both a reactant and a product. Oleum is efficient because it packs more acid into a smaller volume by integrating \( \mathrm{SO}_3 \) into the sulfuric acid solution.

When extra water is added to oleum, the existing \( \mathrm{SO}_3 \) reacts to increase the concentration of sulfuric acid. Thus, the description of oleum’s concentration — such as "104.5% sulfuric acid" — refers to the potential \( \mathrm{H}_2\mathrm{SO}_4 \) capable of being generated upon dilution with water.

Characteristics of sulfuric acid include:

• Strong acidity and highly corrosive.
• Mixes exothermically with water, requiring dilution to produce safer concentrations.
• Vital in chemical manufacturing, metal processing, and automotive applications.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry involves the calculation of reactants and products in chemical reactions. Key to mastering stoichiometry is understanding the mole concept and balancing chemical equations. In the case of oleum chemistry, we look closely at the stoichiometric relationship between \( \mathrm{SO}_3 \) and \( \mathrm{H}_2\mathrm{SO}_4 \).

Let’s consider the relevant reaction:\[ \mathrm{SO}_3 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{H}_2\mathrm{SO}_4 \]This shows a 1:1 molar ratio between \( \mathrm{SO}_3 \) and water to form sulfuric acid. Understanding the molecular weights of those compounds helps in determining how much \( \mathrm{SO}_3 \) contributes to the final mass of \( \mathrm{H}_2\mathrm{SO}_4 \). For instance, with molecular weights of around 80 g/mol for \( \mathrm{SO}_3 \) and 98 g/mol for \( \mathrm{H}_2\mathrm{SO}_4 \), you can calculate the theoretical yield of the reactions.

Important aspects of stoichiometry include:

• Balancing chemical equations based on mass conservation.
• Calculating moles, masses, and percentages of reactants and products.
• Using stoichiometric coefficients to determine the mass relationships in reactions.

Understanding the stoichiometry for sulfur trioxide reactions is essential for predicting the formation of sulfuric acid accurately.