At room temperature, the following reaction proceeds nearly to completion:
`2NO+O_(2)to2NO_(2)toN_(2)O_(4)`
The dimer, `N_(2)O_(4)`, solidfies at `262 K`. A `250 mL` flask and a `100 mL` flask are separated by a stopcock. At `300 K`, the nitric oxide in the larger flask exerts a pressure of `1.053 atm` and the smaller one contains oxygen at `0.789 atm`. The gase are mixed by opening the stopcock and after the end of the reaction the flasks are cooled to `220 K`. Neglecting the vapour pressure of the dimer, find out the pressure and composition of the gas remaining at `220 K`. (Assume the gases to behave ideally)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the moles of NO in the 250 mL flask.
Using the ideal gas equation \( PV = nRT \), we can rearrange it to find the number of moles \( n \):

\[ n = \frac{PV}{RT} \]

For the 250 mL flask:
- Pressure \( P = 1.053 \, \text{atm} \)
- Volume \( V = 250 \, \text{mL} = 0.250 \, \text{L} \)
- Gas constant \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- Temperature \( T = 300 \, \text{K} \)

Substituting the values:

\[ n_{NO} = \frac{(1.053 \, \text{atm})(0.250 \, \text{L})}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(300 \, \text{K})} \]

Calculating this gives:

\[ n_{NO} = \frac{0.26325}{24.63} \approx 0.0107 \, \text{moles} \]

### Step 2: Calculate the moles of O2 in the 100 mL flask.
For the 100 mL flask:
- Pressure \( P = 0.789 \, \text{atm} \)
- Volume \( V = 100 \, \text{mL} = 0.100 \, \text{L} \)

Substituting the values into the ideal gas equation:

\[ n_{O_2} = \frac{(0.789 \, \text{atm})(0.100 \, \text{L})}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(300 \, \text{K})} \]

Calculating this gives:

\[ n_{O_2} = \frac{0.0789}{24.63} \approx 0.0032 \, \text{moles} \]

### Step 3: Determine the limiting reactant and the amount of products formed.
From the reaction \( 2NO + O_2 \rightarrow 2NO_2 \), we can see that:
- 1 mole of \( O_2 \) reacts with 2 moles of \( NO \).

Calculating the moles of \( NO \) required for the available \( O_2 \):

\[ \text{Moles of } NO \text{ required} = 2 \times n_{O_2} = 2 \times 0.0032 \approx 0.0064 \, \text{moles} \]

Since we have \( 0.0107 \, \text{moles} \) of \( NO \), \( O_2 \) is the limiting reactant.

### Step 4: Calculate the moles of \( NO \) remaining after the reaction.
The moles of \( NO \) that react:

\[ \text{Moles of } NO \text{ that react} = 2 \times n_{O_2} = 0.0064 \, \text{moles} \]

Remaining moles of \( NO \):

\[ n_{NO \, \text{remaining}} = n_{NO} - \text{Moles of } NO \text{ that react} = 0.0107 - 0.0064 \approx 0.0043 \, \text{moles} \]

### Step 5: Calculate the moles of \( NO_2 \) produced.
From the stoichiometry of the reaction, the moles of \( NO_2 \) produced will be equal to the moles of \( O_2 \) reacted:

\[ n_{NO_2} = 2 \times n_{O_2} = 0.0064 \, \text{moles} \]

### Step 6: Calculate the total moles of gas remaining at 220 K.
At 220 K, the dimer \( N_2O_4 \) solidifies, so we only consider the remaining \( NO \) and the produced \( NO_2 \):

\[ n_{total} = n_{NO \, \text{remaining}} + n_{NO_2} = 0.0043 + 0.0064 = 0.0107 \, \text{moles} \]

### Step 7: Calculate the pressure at 220 K.
Using the ideal gas equation again:

\[ P = \frac{nRT}{V} \]

The total volume \( V \) after mixing is \( 250 \, \text{mL} + 100 \, \text{mL} = 350 \, \text{mL} = 0.350 \, \text{L} \).

Substituting the values:

\[ P = \frac{(0.0107 \, \text{moles})(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(220 \, \text{K})}{0.350 \, \text{L}} \]

Calculating this gives:

\[ P \approx \frac{0.1945}{0.350} \approx 0.555 \, \text{atm} \]

### Final Answer:
The pressure of the gas remaining at 220 K is approximately **0.555 atm**.