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

Step-by-Step Solution

Answer

The percentage of free \( \mathrm{SO}_{3} \) is 20%.

Step 1: Understanding the Problem

We need to calculate the percentage of free \( \mathrm{SO}_{3} \) in oleum labeled as '104.5% \( \mathrm{H}_{2} \mathrm{SO}_{4} \)'. According to the label, when \( 100 \, \mathrm{g} \) of oleum is diluted, the total mass of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) formed is \( 104.5 \, \mathrm{g} \). We will determine how much of this mass comes from \( \mathrm{SO}_{3} \) reacting with water.

Step 2: Identify Contributions to Mass

The initial \( 100 \, \mathrm{g} \) of oleum contains some mass as \( \mathrm{H}_{2} \mathrm{SO}_{4} \) and some as free \( \mathrm{SO}_{3} \). Free \( \mathrm{SO}_{3} \) reacts with water to form additional \( \mathrm{H}_{2} \mathrm{SO}_{4} \). The total mass after dilution includes the initial mass and the mass added by converting \( \mathrm{SO}_{3} \) to \( \mathrm{H}_{2} \mathrm{SO}_{4} \).

Step 3: Equation Setup

When free \( \mathrm{SO}_{3} \) (\( x \) grams) combines with water, each mole of \( \mathrm{SO}_{3} \) (80 g/mol) becomes \( \mathrm{H}_{2} \mathrm{SO}_{4} \) (98 g/mol). This means \( x \) grams of free \( \mathrm{SO}_{3} \) when converted forms \( \frac{98}{80}x \) grams of \( \mathrm{H}_{2} \mathrm{SO}_{4} \).

Step 4: Calculating Free SO3 Percentage

In \( 104.5 \, \mathrm{g} \) of \( \mathrm{H}_{2} \mathrm{SO}_{4} \), \( 100 \, \mathrm{g} - x \) is the original \( \mathrm{H}_{2} \mathrm{SO}_{4} \) and the rest is from \( \mathrm{SO}_{3} \). Thus, \( 100 - x + \frac{98}{80}x = 104.5 \). Solve this equation for \( x \).

Step 5: Solve the Equation

Simplify and solve: \( 100 - x + 1.225x = 104.5 \). This simplifies to \( 0.225x = 4.5 \), yielding \( x = \frac{4.5}{0.225} = 20 \). So, \( 20 \) grams of \( SO_3 \) were initially present in the oleum.

Key Concepts

Sulfur trioxide (SO3)

Sulfur trioxide (SO₃) is a significant chemical primarily in the industrial production of sulfuric acid. It is a simple molecule with one sulfur atom bonded to three oxygen atoms. SO₃ is formed from the oxidation of sulfur dioxide in a contact process.
In its gaseous phase, SO₃ exists as a monomer, while in liquid form, it assumes a polymeric form. Due to its high reactivity, it is generally handled with caution, often dissolved in sulfuric acid, creating oleum.
When SO₃ comes into contact with water, it rapidly forms sulfuric acid ( H₂SO₄), releasing a significant amount of heat in an exothermic reaction:

• SO₃ + H₂O → H₂SO₄

Understanding SO₃'s role is essential in industrial applications and chemistry, especially for those working with oleum, where SO₃'s behavior impacts the concentration and properties of resulting solutions.

Sulfuric acid (H2SO4)

Sulfuric acid (H₂SO₄) is one of the most widely used chemicals in various industries. Its primary use is in the manufacture of fertilizers but is also crucial in petroleum refining, wastewater processing, and chemical synthesis. This compound is a dense, oily liquid that is soluble in water, releasing a lot of heat upon dissolution.
H₂SO₄ is called the "king of chemicals" because of its wide-ranging applications and usefulness. In oleum, which is a mixture of H₂SO₄ and excess SO₃, sulfuric acid plays a pivotal role as both a reactant and solvent.
Oleum is labeled according to the amount of pure H₂SO₄ produced when the solution is diluted with water. This labeling indicates the strength of the oleum solution, showing how much free SO₃ it contains and how much additional H₂SO₄ will form upon interaction with water.
The ability of H₂SO₄ to act as an acid, dehydrant, oxidizing agent, and reactant with many other compounds underscores its versatile nature.

Dilution process calculation

The dilution process in oleum chemistry involves calculating the percentage of free SO₃ that can potentially convert into additional sulfuric acid. When you encounter an oleum labeled as a certain percentage of H₂SO₄, it describes the total mass of H₂SO₄ that will result from diluting 100g of that oleum with water.
To determine how much SO₃ is free in the oleum, we set up an equation based on the chemical reaction where SO₃ reacts with water to form H₂SO₄. For example, if oleum is labeled as 104.5% H₂SO₄, when diluted, it yields 104.5g total mass of sulfuric acid from 100g of the original oleum, including any new H₂SO₄ formed by the reaction with free SO₃.
The formula

• 100 - x + (\( \frac{98}{80} \) \( x \) ) = 104.5

is used to calculate the free SO₃ content, where \( x \) represents grams of free SO₃. Solving this equation gives you the mass of free SO₃, which is essential in determining the concentration and maximum yield of H₂SO₄ that could be achieved when the oleum is fully diluted.