At `27^(@)C` NO and `Cl_(2)` gases are introduced in a 10 litre flask such that their initial partial pressures are 20 and 16 atm respectively. The flask already contains 24 g of magnesium. After some time, the amount of magnesium left was 0.2 moles due to the establishment of following two equilibria
`2NO(g)+Cl_(2)(g) hArr 2NOCl(g)`
`Mg(s)+Cl_(2)(g) hArr 2MgCl_(2)(s), K_(p)=0.2 "atm"^(-1)`
The final pressure of NOCl would be

Options

A. 7.84 atm
B. 18.06 atm
C. 129.6 atm
D. 64.8 atm

Explanation

To solve the problem, we need to analyze the two equilibria involving NO, Cl₂, and Mg, and then calculate the final pressure of NOCl in the system.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Initial partial pressures:
- \( P_{NO} = 20 \, \text{atm} \)
- \( P_{Cl_2} = 16 \, \text{atm} \)
- The volume of the flask is \( 10 \, \text{L} \).
- The initial amount of magnesium is \( 24 \, \text{g} \).

2. **Convert Mass of Magnesium to Moles:**
- Molar mass of magnesium (Mg) = 24 g/mol.
- Initial moles of Mg = \( \frac{24 \, \text{g}}{24 \, \text{g/mol}} = 1 \, \text{mol} \).

3. **Determine Moles of Magnesium Left:**
- After some time, the amount of magnesium left is \( 0.2 \, \text{mol} \).
- Therefore, moles of magnesium reacted = \( 1 - 0.2 = 0.8 \, \text{mol} \).

4. **Establish the First Equilibrium Reaction:**
- The first reaction is:
\[ 2NO(g) + Cl_2(g) \rightleftharpoons 2NOCl(g) \]
- Let \( x \) be the change in moles of NO that reacts. Then:
- Change in moles of \( Cl_2 \) = \( \frac{x}{2} \)
- Change in moles of \( NOCl \) = \( x \)

5. **Set Up the Equilibrium Expression:**
- Initial pressures:
- \( P_{NO} = 20 \, \text{atm} \)
- \( P_{Cl_2} = 16 \, \text{atm} \)
- \( P_{NOCl} = 0 \, \text{atm} \)
- At equilibrium:
- \( P_{NO} = 20 - 2x \)
- \( P_{Cl_2} = 16 - x \)
- \( P_{NOCl} = 2x \)

6. **Use the Second Equilibrium Reaction:**
- The second reaction is:
\[ Mg(s) + Cl_2(g) \rightleftharpoons 2MgCl_2(s) \]
- The equilibrium constant \( K_p \) for this reaction is given as \( 0.2 \, \text{atm}^{-1} \).
- The equilibrium expression is:
\[ K_p = \frac{(P_{MgCl_2})^2}{P_{Cl_2}} = 0.2 \]
- Since \( MgCl_2 \) is a solid, its pressure is not included in the expression.

7. **Calculate the Partial Pressure of \( Cl_2 \):**
- From the reaction, we know \( 0.8 \, \text{mol} \) of \( Cl_2 \) reacts with \( 0.8 \, \text{mol} \) of magnesium.
- Thus, \( P_{Cl_2} = 16 - 0.8 = 15.2 \, \text{atm} \).

8. **Substituting into the Equilibrium Expression:**
- Rearranging the \( K_p \) expression:
\[ 0.2 = \frac{(0)^2}{P_{Cl_2}} \implies P_{Cl_2} = 5 \, \text{atm} \]

9. **Calculate the Value of \( x \):**
- From the equilibrium of the first reaction:
\[ 16 - x = 5 \implies x = 11 \, \text{atm} \]

10. **Final Calculation of \( P_{NOCl} \):**
- Since \( P_{NOCl} = 2x \):
\[ P_{NOCl} = 2 \times 11 = 22 \, \text{atm} \]

### Final Answer:
The final pressure of NOCl is **22 atm**.