At `250^@C`, `K_c` for `PCl_5(g)hArr PCl_3(g)+Cl_2(g)` is `0.04`. How many moles of `PCl_5` must be added to a `3-L` flask to obtain `0.15M Cl_2` at equilibrium?

Options

A. `2.1 mol`
B. `1.7mol`
C. `0.9 mol`
D. `3.5 mol`

Explanation

To solve the problem step by step, we will analyze the equilibrium reaction and use the given equilibrium constant to find the required number of moles of \( PCl_5 \) to achieve the desired concentration of \( Cl_2 \).

### Step 1: Write the balanced equation
The balanced chemical equation for the dissociation of \( PCl_5 \) is:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

### Step 2: Define initial conditions
Let \( A \) be the initial number of moles of \( PCl_5 \) added to the flask. The volume of the flask is given as \( 3 \, L \).

### Step 3: Define changes at equilibrium
Let \( X \) be the number of moles of \( PCl_5 \) that dissociate at equilibrium. At equilibrium, the moles of each species will be:
- Moles of \( PCl_5 \): \( A - X \)
- Moles of \( PCl_3 \): \( X \)
- Moles of \( Cl_2 \): \( X \)

### Step 4: Use the given concentration of \( Cl_2 \)
We are given that the equilibrium concentration of \( Cl_2 \) is \( 0.15 \, M \). Since \( Cl_2 \) is produced in a 1:1 ratio with \( PCl_5 \), we have:
\[ X = 0.15 \, \text{moles/L} \times 3 \, L = 0.45 \, \text{moles} \]

### Step 5: Write the expression for the equilibrium constant \( K_c \)
The expression for the equilibrium constant \( K_c \) for the reaction is:
\[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \]
Substituting the equilibrium concentrations, we have:
\[ K_c = \frac{(0.15)(0.15)}{A - 0.45} \]
Given that \( K_c = 0.04 \), we can set up the equation:
\[ 0.04 = \frac{(0.15)(0.15)}{A - 0.45} \]

### Step 6: Solve for \( A \)
Rearranging the equation gives:
\[ 0.04(A - 0.45) = 0.0225 \]
\[ 0.04A - 0.018 = 0.0225 \]
\[ 0.04A = 0.0225 + 0.018 \]
\[ 0.04A = 0.0405 \]
\[ A = \frac{0.0405}{0.04} = 1.0125 \, \text{moles} \]

### Step 7: Calculate the total moles of \( PCl_5 \)
To find the total moles of \( PCl_5 \) required, we multiply the concentration by the volume:
\[ \text{Total moles of } PCl_5 = A \times \text{Volume} = 1.0125 \, \text{moles} \times 3 \, L = 3.0375 \, \text{moles} \]

### Step 8: Final answer
The number of moles of \( PCl_5 \) that must be added to the flask to obtain \( 0.15 \, M \) \( Cl_2 \) at equilibrium is approximately:
\[ \boxed{3.04} \, \text{moles} \]