In 109% `H_(2)SO_(4)` labelled oleum, the percent of free `SO_(3)` and `H_(2)SO_(4)` are

Options

A. 30%, 70%
B. 40%, 60%
C. 60%, 40%
D. 15%, 85%

Answer

To solve the problem of determining the percentage of free \( SO_3 \) and \( H_2SO_4 \) in 109% \( H_2SO_4 \) labelled oleum, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Oleum**:
Oleum is a mixture of sulfuric acid (\( H_2SO_4 \)) and sulfur trioxide (\( SO_3 \)). The formula for oleum can be expressed as:
\[ \text{Oleum} = H_2SO_4 + SO_3 \]

2. **Interpreting 109% Oleum**:
The term "109% \( H_2SO_4 \) labelled oleum" means that when 100 grams of this oleum is diluted with water, it produces 109 grams of pure sulfuric acid. This indicates that the total mass of sulfuric acid produced is 109 grams.

3. **Calculating the Mass of Water**:
When 100 grams of oleum is mixed with 9 grams of water, the water reacts with the free \( SO_3 \) to form additional \( H_2SO_4 \). The total mass of the solution after dilution is:
\[ 100 \text{ g (oleum)} + 9 \text{ g (water)} = 109 \text{ g} \]

4. **Determining the Mass of Free \( SO_3 \)**:
The amount of water used (9 grams) reacts with \( SO_3 \) to form \( H_2SO_4 \). The reaction can be represented as:
\[ H_2O + SO_3 \rightarrow H_2SO_4 \]
From the molar mass of water (18 g/mol), we can find out how much \( SO_3 \) reacts with 9 grams of water:
\[ \text{Moles of } H_2O = \frac{9 \text{ g}}{18 \text{ g/mol}} = 0.5 \text{ moles} \]
Since 1 mole of \( H_2O \) reacts with 1 mole of \( SO_3 \), 0.5 moles of \( H_2O \) will react with 0.5 moles of \( SO_3 \).

5. **Calculating the Mass of \( SO_3 \)**:
The molar mass of \( SO_3 \) is approximately 80 g/mol. Therefore, the mass of \( SO_3 \) that reacts is:
\[ \text{Mass of } SO_3 = 0.5 \text{ moles} \times 80 \text{ g/mol} = 40 \text{ g} \]

6. **Finding the Percentage of Free \( SO_3 \)**:
Since we have 40 grams of \( SO_3 \) in 100 grams of oleum, the percentage of free \( SO_3 \) is:
\[ \text{Percentage of } SO_3 = \left( \frac{40 \text{ g}}{100 \text{ g}} \right) \times 100 = 40\% \]

7. **Finding the Percentage of Free \( H_2SO_4 \)**:
The total mass of oleum is 100 grams, and since 40 grams is \( SO_3 \), the remaining mass must be \( H_2SO_4 \):
\[ \text{Mass of } H_2SO_4 = 100 \text{ g} - 40 \text{ g} = 60 \text{ g} \]
Therefore, the percentage of free \( H_2SO_4 \) is:
\[ \text{Percentage of } H_2SO_4 = \left( \frac{60 \text{ g}}{100 \text{ g}} \right) \times 100 = 60\% \]

### Final Answer:
- Percentage of free \( SO_3 \) = 40%
- Percentage of free \( H_2SO_4 \) = 60%