Oleum is considered as a solution of `SO_(3)` in `H_(2)SO_(4)`, which is obtained by passing `SO_(3)` in solution of `H_(2)SO_(4)` When 100 g sample of oleum is diluted with desired mass of `H_(2)O` then the total mass of `H_(2)SO_(4)` obtained after dilution is known is known as % labelling in oleum.
For example, a oleum bottle labelled as '`109% H_(2)SO_(4)`' means the 109 g total mass of pure `H_(2)SO_(4)` will be formed when 100 g of oleum is diluted by 9 g of `H_(2)O` which combines with all the free `SO_(3)` present in oleum to form `H_(2)SO_(4)` as `SO_(3)+H_(2)O to H_(2)SO_(4)`
If excess water is added into a bottle sample labelled as "112%`H_(2)SO_(4)`" and is reacted with 5.3 g `NaCO_(3)` then find the volume of `CO_(2)` evolved at 1 atm pressure and 300 K temperature after the completion of the reaction :

Options

A. 2.46 L
B. 24.6 L
C. 1.23 L
D. 12.3 L

Answer

To solve the problem step-by-step, we will follow the outlined procedure:

### Step 1: Understand the composition of oleum
Oleum is a solution of \( SO_3 \) in \( H_2SO_4 \). The percentage labeling indicates the total mass of \( H_2SO_4 \) that can be obtained from a given mass of oleum when diluted with water.

### Step 2: Calculate the mass of \( H_2SO_4 \) from the oleum
Given that the oleum is labeled as "112% \( H_2SO_4 \)", it means that when 100 g of oleum is diluted, it will yield 112 g of pure \( H_2SO_4 \).

### Step 3: Write the reaction between \( Na_2CO_3 \) and \( H_2SO_4 \)
The balanced chemical reaction is:
\[ Na_2CO_3 + H_2SO_4 \rightarrow Na_2SO_4 + CO_2 + H_2O \]
From this equation, we can see that 1 mole of \( Na_2CO_3 \) reacts with 1 mole of \( H_2SO_4 \) to produce 1 mole of \( CO_2 \).

### Step 4: Calculate the moles of \( Na_2CO_3 \)
We are given 5.3 g of \( Na_2CO_3 \). To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
The molar mass of \( Na_2CO_3 \) is approximately 106 g/mol. Therefore:
\[ \text{Number of moles of } Na_2CO_3 = \frac{5.3 \, \text{g}}{106 \, \text{g/mol}} \approx 0.050 \, \text{moles} \]

### Step 5: Determine the moles of \( CO_2 \) produced
Since the reaction shows a 1:1 ratio between \( Na_2CO_3 \) and \( CO_2 \), the moles of \( CO_2 \) produced will also be 0.050 moles.

### Step 6: Use the ideal gas equation to find the volume of \( CO_2 \)
The ideal gas equation is:
\[ PV = nRT \]
Rearranging for volume \( V \):
\[ V = \frac{nRT}{P} \]
Where:
- \( n = 0.050 \, \text{moles} \)
- \( R = 0.0821 \, \text{L atm/(mol K)} \)
- \( T = 300 \, \text{K} \)
- \( P = 1 \, \text{atm} \)

Substituting the values:
\[ V = \frac{(0.050 \, \text{moles}) \times (0.0821 \, \text{L atm/(mol K)}) \times (300 \, \text{K})}{1 \, \text{atm}} \]
Calculating:
\[ V \approx \frac{1.2315 \, \text{L atm}}{1 \, \text{atm}} \approx 1.23 \, \text{L} \]

### Final Answer
The volume of \( CO_2 \) evolved at 1 atm pressure and 300 K temperature is approximately **1.23 liters**.

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