Sulphur can form dihalide, tetrahalide and hexahalide with fluorine. One mole of each of these three compounds is mixed with water. The total number of moles of product molecules obtained is ____. If no reaction occurs, count zero

Answer

To solve the problem, we need to analyze the reactions of sulfur halides with water and count the total number of moles of product molecules formed.

### Step-by-Step Solution:

1. **Identify the Sulfur Halides**:
- The dihalide is \( SF_2 \).
- The tetrahalide is \( SF_4 \).
- The hexahalide is \( SF_6 \).

2. **Reaction of \( SF_2 \) with Water**:
- The reaction can be represented as:
\[ SF_2 + 2H_2O \rightarrow 2HF + SO_2 \]
- From this reaction, we see that:
- 1 mole of \( SF_2 \) produces 2 moles of \( HF \) and 1 mole of \( SO_2 \).
- Total moles of products = \( 2 + 1 = 3 \) moles.

3. **Reaction of \( SF_4 \) with Water**:
- The reaction can be represented as:
\[ SF_4 + 2H_2O \rightarrow 4HF + H_2SO3 \]
- From this reaction, we see that:
- 1 mole of \( SF_4 \) produces 4 moles of \( HF \) and 1 mole of \( H_2SO3 \).
- Total moles of products = \( 4 + 1 = 5 \) moles.

4. **Reaction of \( SF_6 \) with Water**:
- The reaction can be represented as:
\[ SF_6 + 3H_2O \rightarrow 6HF + SO_3 \]
- From this reaction, we see that:
- 1 mole of \( SF_6 \) produces 6 moles of \( HF \) and 1 mole of \( SO_3 \).
- Total moles of products = \( 6 + 1 = 7 \) moles.

5. **Calculate Total Moles of Products**:
- Now, we sum the total moles of products from each reaction:
- From \( SF_2 \): 3 moles
- From \( SF_4 \): 5 moles
- From \( SF_6 \): 7 moles
- Total = \( 3 + 5 + 7 = 15 \) moles.

### Final Answer:
The total number of moles of product molecules obtained is **15**.