A flask of `4.48 L` capacity contains a mixture of `N_(2)` and `H_(2)` at `0^(@)C` and `1 atm` pressure. If thed mixture is made to react to form `NH_(3)` gas at the same temperature, the pressure in the flask reduces to `0.75 atm`. Select statement(s) :

Options

A. Initially total moves of gases in the mixture is `0.2 mol`.
B. Initially total moves of gases in the mixture is `0.4 mol`.
C. The partial pressure of `NH_(3)` gas in the final mixture is `0.33 atm`.
D. The partial pressure of `NH_(3)` gas in the final mixture is `0.25 atm`.

Explanation

To solve the problem step by step, we will analyze the reaction and the changes in pressure and moles of gases involved.

### Step 1: Write the balanced chemical equation
The reaction between nitrogen and hydrogen to form ammonia is given by:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 2: Define initial conditions
We know the total pressure of the gas mixture (N₂ and H₂) is 1 atm. Let:
- The partial pressure of \( N_2 \) be \( x \) atm
- The partial pressure of \( H_2 \) be \( y \) atm

From the problem, we have:
\[ x + y = 1 \quad \text{(1)} \]

### Step 3: Define changes during the reaction
Let \( P \) be the change in pressure due to the reaction. According to the stoichiometry of the reaction:
- For every 1 mole of \( N_2 \) that reacts, 3 moles of \( H_2 \) react to produce 2 moles of \( NH_3 \).

Thus, if \( P \) atm of \( N_2 \) reacts, then \( 3P \) atm of \( H_2 \) reacts, and \( 2P \) atm of \( NH_3 \) is produced.

### Step 4: Write the final pressures
After the reaction, the pressures will be:
- Pressure of \( N_2 \): \( x - P \)
- Pressure of \( H_2 \): \( y - 3P \)
- Pressure of \( NH_3 \): \( 2P \)

The total pressure after the reaction is given as 0.75 atm:
\[ (x - P) + (y - 3P) + 2P = 0.75 \quad \text{(2)} \]

### Step 5: Substitute equation (1) into equation (2)
From equation (1), we can substitute \( y = 1 - x \) into equation (2):
\[ (x - P) + ((1 - x) - 3P) + 2P = 0.75 \]
This simplifies to:
\[ 1 - 2P = 0.75 \]

### Step 6: Solve for \( P \)
Rearranging gives:
\[ 2P = 1 - 0.75 \]
\[ 2P = 0.25 \]
\[ P = 0.125 \]

### Step 7: Calculate the pressure of \( NH_3 \)
The pressure of \( NH_3 \) produced is:
\[ 2P = 2 \times 0.125 = 0.25 \text{ atm} \]

### Step 8: Calculate initial moles of gases
Using the ideal gas law, we can find the number of moles of gases initially present:
\[ PV = nRT \]
Where:
- \( P = 1 \text{ atm} \)
- \( V = 4.48 \text{ L} \)
- \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \) (ideal gas constant)
- \( T = 273 \text{ K} \) (0°C)

Calculating the initial moles:
\[ n = \frac{PV}{RT} = \frac{(1 \text{ atm})(4.48 \text{ L})}{(0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1})(273 \text{ K})} \]
\[ n \approx 0.2 \text{ moles} \]

### Summary of Results
- The initial total moles of gases in the mixture is approximately 0.2 moles.
- The pressure of ammonia gas in the final mixture is 0.25 atm.