To solve the problem, we need to determine the amount of free \( SO_3 \) remaining in the solution after adding 9 g of water to the oleum sample labeled as "112% \( H_2SO_4 \)".
### Step-by-Step Solution:
1. **Understanding the Labeling**:
- The oleum is labeled as "112% \( H_2SO_4 \)", which means that when 100 g of oleum is diluted, it will yield a total of 112 g of \( H_2SO_4 \).
- This implies that the mass of free \( SO_3 \) in the oleum is \( 112 \, \text{g} - 100 \, \text{g} = 12 \, \text{g} \).
2. **Determine the Moles of Free \( SO_3 \)**:
- The molar mass of \( SO_3 \) is approximately \( 80 \, \text{g/mol} \) (32 for S + 3 × 16 for O).
- Calculate the moles of \( SO_3 \):
\[
\text{Moles of } SO_3 = \frac{12 \, \text{g}}{80 \, \text{g/mol}} = 0.15 \, \text{mol}
\]
3. **Water Added**:
- 9 g of water is added to the oleum.
- The molar mass of water \( H_2O \) is approximately \( 18 \, \text{g/mol} \).
- Calculate the moles of water added:
\[
\text{Moles of } H_2O = \frac{9 \, \text{g}}{18 \, \text{g/mol}} = 0.5 \, \text{mol}
\]
4. **Reaction Between Water and \( SO_3 \)**:
- The reaction is:
\[
H_2O + SO_3 \rightarrow H_2SO_4
\]
- From stoichiometry, 1 mole of \( H_2O \) reacts with 1 mole of \( SO_3 \).
- Therefore, 0.5 moles of \( H_2O \) will react with 0.5 moles of \( SO_3 \).
5. **Calculate Remaining \( SO_3 \)**:
- Initial moles of \( SO_3 \) = 0.15 mol
- Moles of \( SO_3 \) that reacted = 0.5 mol
- Since we only have 0.15 mol of \( SO_3 \), all of it will react, and there will be no \( SO_3 \) remaining after the reaction:
\[
\text{Remaining } SO_3 = 0.15 \, \text{mol} - 0.15 \, \text{mol} = 0 \, \text{mol}
\]
6. **Conclusion**:
- After adding 9 g of water, all the free \( SO_3 \) will have reacted, leaving us with 0 moles of free \( SO_3 \).
### Final Answer:
The amount of free \( SO_3 \) remaining in the solution is **0 g**.
