30 gm `H_(2)SO_(4)` is mixed with 20 gram `SO_(3)` to form mixture
Determine % labelling of oleum solution

Options

A. `104.5`
B. 106
C. 109
D. 110

Answer

To determine the percent labeling of the oleum solution formed by mixing 30 grams of \( H_2SO_4 \) with 20 grams of \( SO_3 \), we can follow these steps:

### Step 1: Understanding the Reaction
The reaction between sulfur trioxide (\( SO_3 \)) and water (\( H_2O \)) produces sulfuric acid (\( H_2SO_4 \)):
\[ SO_3 + H_2O \rightarrow H_2SO_4 \]
In this reaction, 1 mole of \( SO_3 \) reacts with 1 mole of water to produce 1 mole of \( H_2SO_4 \).

### Step 2: Calculate Molar Masses
- Molar mass of \( H_2SO_4 \) = 2 (H) + 32 (S) + 4(16) (O) = 98 g/mol
- Molar mass of \( SO_3 \) = 32 (S) + 3(16) (O) = 80 g/mol
- Molar mass of \( H_2O \) = 2 (H) + 16 (O) = 18 g/mol

### Step 3: Determine Moles of Reactants
- Moles of \( H_2SO_4 \):
\[ \text{Moles of } H_2SO_4 = \frac{30 \text{ g}}{98 \text{ g/mol}} \approx 0.306 \text{ moles} \]
- Moles of \( SO_3 \):
\[ \text{Moles of } SO_3 = \frac{20 \text{ g}}{80 \text{ g/mol}} = 0.25 \text{ moles} \]

### Step 4: Identify Limiting Reactant
Since \( SO_3 \) is present in a smaller amount (0.25 moles), it is the limiting reactant.

### Step 5: Calculate Amount of Water Required
According to the stoichiometry of the reaction, 1 mole of \( SO_3 \) requires 1 mole of \( H_2O \). Therefore, for 0.25 moles of \( SO_3 \), we need:
\[ \text{Water required} = 0.25 \text{ moles} \times 18 \text{ g/mol} = 4.5 \text{ g} \]

### Step 6: Calculate Total Mass of the Oleum Solution
The total mass of the oleum solution is the mass of \( H_2SO_4 \) plus the mass of \( SO_3 \):
\[ \text{Total mass} = 30 \text{ g } H_2SO_4 + 20 \text{ g } SO_3 = 50 \text{ g} \]

### Step 7: Calculate Percent Labeling
The percent labeling of oleum is given by:
\[ \text{Percent labeling} = 100 + X \]
where \( X \) is the mass of water required. Therefore:
\[ \text{Percent labeling} = 100 + 4.5 = 104.5\% \]

### Final Answer
The percent labeling of the oleum solution is **104.5%**.

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