Oleum is mixture of `H_(2)SO_(4)` and `SO_(3)` i.e. `H_(2)S_(2)O_(7)` which is obtained by passing `SO_(3)` is solution of `H_(2)SO_(4)`. In order to dissolve `SO_(3)` in oleum, dilution of oleum is done by water in which oleum is converted into pure `H_(2)SO` as shown below:
`H_(2)SO_(4)+SO_(3)+H_(2)Oto2H_(2)SO_(4)` (pure)
When 100 gm oleum is diluted with water then total mass of diluted oleum is known as percentage labelling in oleum.
For example: `109% H_(2)SO_(4)` labelling of oleum sample means that 109 gm pure `H_(2)SO_(4)` is obtained on diluting 100 gm oleum with 9 gm `H_(2)O` which dissolves al free `SO_(3)` in oleum.
If the number of moles of free `SO_(3), H_(2)SO_(4)`, and `H_(2)O` be x, y and z respectively in 118%` H_(2)SO_(4)` labelled oleum, the value of `(x+y+z)` is

Options

A. 2.2
B. 3.2
C. 3.4
D. 4.2

Answer

To solve the problem step by step, we will analyze the given information about oleum and the chemical reaction involved in diluting it with water.

### Step 1: Understanding the Composition of Oleum
Oleum is a mixture of sulfuric acid (H₂SO₄) and sulfur trioxide (SO₃). The reaction when oleum is diluted with water can be represented as:
\[ \text{H}_2\text{SO}_4 + \text{SO}_3 + \text{H}_2\text{O} \rightarrow 2 \text{H}_2\text{SO}_4 \]

### Step 2: Analyzing the Given Information
We are given that the oleum is labeled as 118% H₂SO₄, which means that for every 100 grams of oleum, we can obtain 118 grams of pure H₂SO₄ after dilution.

### Step 3: Calculating the Mass of Water
From the example provided in the question, if 100 grams of oleum yields 118 grams of pure H₂SO₄, we can determine the mass of water added:
- Mass of oleum = 100 g
- Mass of pure H₂SO₄ obtained = 118 g
- Therefore, the mass of water added = 118 g - 100 g = 18 g

### Step 4: Finding the Moles of H₂SO₄
The molar mass of H₂SO₄ is 98 g/mol. Thus, the number of moles of H₂SO₄ produced can be calculated as follows:
\[ \text{Moles of H}_2\text{SO}_4 = \frac{118 \text{ g}}{98 \text{ g/mol}} \approx 1.20 \text{ moles} \]

### Step 5: Finding the Moles of Oleum
The molar mass of oleum (H₂S₂O₇) can be calculated as follows:
- H₂S₂O₇ = 2(1) + 2(32) + 7(16) = 178 g/mol
Thus, the number of moles of oleum in 100 g is:
\[ \text{Moles of oleum} = \frac{100 \text{ g}}{178 \text{ g/mol}} \approx 0.56 \text{ moles} \]

### Step 6: Finding the Moles of SO₃
From the balanced equation, 1 mole of oleum produces 0.5 moles of SO₃. Therefore, the number of moles of SO₃ can be calculated as:
\[ \text{Moles of SO}_3 = 0.56 \text{ moles of oleum} \times \frac{1 \text{ mole of SO}_3}{2 \text{ moles of oleum}} = 0.28 \text{ moles} \]

### Step 7: Finding the Moles of Water
The number of moles of water (H₂O) can be calculated using its molar mass (18 g/mol):
\[ \text{Moles of H}_2\text{O} = \frac{18 \text{ g}}{18 \text{ g/mol}} = 1 \text{ mole} \]

### Step 8: Summing the Moles
Now, we can sum the moles of SO₃, H₂SO₄, and H₂O:
\[ x + y + z = \text{Moles of SO}_3 + \text{Moles of H}_2\text{SO}_4 + \text{Moles of H}_2\text{O} \]
\[ x + y + z = 0.28 + 1.20 + 1 = 2.48 \text{ moles} \]

### Final Answer
The value of \( x + y + z \) is approximately 2.48 moles.