`K_C" for "PCl_(5) (g) Leftrightarrow PCl_(3) (g)+Cl_(2)(g)` is 0.04 at `250^@C`. How many mole of `PCl_5` must be added to a 3-litre flask to obtain a `Cl_2` concentration of 0.15M?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction and Given Data
The reaction is:
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]

We are given:
- \( K_c = 0.04 \) at \( 250^\circ C \)
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