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. 5
B. 3
C. 6
D. 4.5

Explanation

To solve the problem, we need to analyze the reaction and the changes in the number of moles of each substance involved. The reaction we are considering is:

\[ \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 \]

### Step 1: Write down the initial moles of each substance.
Initially, we have:
- Moles of PCl5 = 3
- Moles of PCl3 = 3
- Moles of Cl2 = 2

### Step 2: Define the change in moles at equilibrium.
Let \( x \) be the number of moles of PCl5 that decomposes at equilibrium. Therefore, at equilibrium, the moles of each substance will be:
- Moles of PCl5 = \( 3 - x \)
- Moles of PCl3 = \( 3 + x \)
- Moles of Cl2 = \( 2 + x \)

### Step 3: Use the information given in the problem.
According to the problem, at equilibrium, there are 1.5 moles of PCl5. Thus, we can set up the equation:
\[ 3 - x = 1.5 \]

### Step 4: Solve for \( x \).
Rearranging the equation gives:
\[ x = 3 - 1.5 = 1.5 \]

### Step 5: Calculate the moles of PCl3 at equilibrium.
Now that we have \( x \), we can find the number of moles of PCl3 at equilibrium:
\[ \text{Moles of PCl3} = 3 + x = 3 + 1.5 = 4.5 \]

### Final Answer:
The number of moles of PCl3 present at equilibrium is **4.5 moles**.

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