Three moles of `PCl_(5)`, three moles of `PCl_(3)` and two moles of `Cl_(2)` are taken in a closed vessel. If at equilibium, the vessel has 1.5 moles of `PCl_(5)` the number of moles of `PCl_(3)` present in it is

Options

A. 6
B. 4.5
C. 5
D. 3

Explanation

To solve the problem, we need to analyze the equilibrium reaction involving the compounds PCl5, PCl3, and Cl2. The reaction can be represented as follows:

\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]

### Step-by-Step Solution:

1. **Identify Initial Moles:**
- Initially, we have:
- \( n(PCl_5) = 3 \) moles
- \( n(PCl_3) = 3 \) moles
- \( n(Cl_2) = 2 \) moles

2. **Set Up the Change in Moles:**
- Let \( x \) be the amount of \( PCl_5 \) that dissociates at equilibrium.
- Therefore, the changes in moles will be:
- \( PCl_5 \) decreases by \( x \) moles.
- \( PCl_3 \) increases by \( x \) moles.
- \( Cl_2 \) increases by \( x \) moles.

3. **Calculate Moles at Equilibrium:**
- At equilibrium, we know that there are 1.5 moles of \( PCl_5 \):
\[ n(PCl_5) = 3 - x = 1.5 \]
- From this equation, we can solve for \( x \):
\[ 3 - x = 1.5 \implies x = 3 - 1.5 = 1.5 \]

4. **Determine Moles of \( PCl_3 \):**
- Now we can find the moles of \( PCl_3 \) at equilibrium:
\[ n(PCl_3) = 3 + x = 3 + 1.5 = 4.5 \]

5. **Conclusion:**
- Therefore, the number of moles of \( PCl_3 \) present at equilibrium is **4.5 moles**.

### Final Answer:
The number of moles of \( PCl_3 \) present in the vessel at equilibrium is **4.5 moles**.