At STP, a container has 1 mole of He, 2 mole Ne, 3 mole `O_(2)` and 4 mole `N_(2)` without changing total pressure if 2 mole of `O_(2)` is removed. Calculate the percentage decrease in the partial pressure of `O_(2)`

Explanation

To solve the problem, we need to calculate the percentage decrease in the partial pressure of \( O_2 \) after removing 2 moles of it from the container. We will follow these steps:

### Step 1: Calculate the initial total number of moles
Initially, the container has:
- 1 mole of He
- 2 moles of Ne
- 3 moles of \( O_2 \)
- 4 moles of \( N_2 \)

Total moles (\( n_{total} \)) = \( 1 + 2 + 3 + 4 = 10 \) moles.

### Step 2: Calculate the initial partial pressure of \( O_2 \)
The mole fraction of \( O_2 \) (\( X_{O_2} \)) is given by:
\[ X_{O_2} = \frac{\text{moles of } O_2}{\text{total moles}} = \frac{3}{10} \]
The initial partial pressure of \( O_2 \) (\( P_{O_2, initial} \)) can be calculated using the formula:
\[ P_{O_2, initial} = X_{O_2} \times P_{total} = \frac{3}{10} \times P_{total} \]

### Step 3: Calculate the new total number of moles after removing \( O_2 \)
After removing 2 moles of \( O_2 \), the moles of \( O_2 \) left are:
\[ 3 - 2 = 1 \text{ mole of } O_2 \]
The new total number of moles (\( n_{total, new} \)) becomes:
\[ 10 - 2 = 8 \text{ moles} \]

### Step 4: Calculate the new partial pressure of \( O_2 \)
Now, the new mole fraction of \( O_2 \) (\( X_{O_2, new} \)) is:
\[ X_{O_2, new} = \frac{\text{moles of } O_2}{\text{new total moles}} = \frac{1}{8} \]
The new partial pressure of \( O_2 \) (\( P_{O_2, new} \)) is:
\[ P_{O_2, new} = X_{O_2, new} \times P_{total} = \frac{1}{8} \times P_{total} \]

### Step 5: Calculate the decrease in partial pressure of \( O_2 \)
The decrease in partial pressure of \( O_2 \) is:
\[ \Delta P_{O_2} = P_{O_2, initial} - P_{O_2, new} = \left(\frac{3}{10} \times P_{total}\right) - \left(\frac{1}{8} \times P_{total}\right) \]

### Step 6: Find a common denominator and simplify
To subtract these fractions, we need a common denominator. The least common multiple of 10 and 8 is 40.
\[ P_{O_2, initial} = \frac{3}{10} \times P_{total} = \frac{12}{40} \times P_{total} \]
\[ P_{O_2, new} = \frac{1}{8} \times P_{total} = \frac{5}{40} \times P_{total} \]
Thus,
\[ \Delta P_{O_2} = \left(\frac{12}{40} - \frac{5}{40}\right) \times P_{total} = \frac{7}{40} \times P_{total} \]

### Step 7: Calculate the percentage decrease in partial pressure of \( O_2 \)
The percentage decrease in partial pressure is given by:
\[ \text{Percentage decrease} = \left(\frac{\Delta P_{O_2}}{P_{O_2, initial}}\right) \times 100 \]
Substituting the values:
\[ \text{Percentage decrease} = \left(\frac{\frac{7}{40} \times P_{total}}{\frac{12}{40} \times P_{total}}\right) \times 100 = \left(\frac{7}{12}\right) \times 100 \]
Calculating this gives:
\[ \text{Percentage decrease} = \frac{7 \times 100}{12} = 58.33\% \]

### Final Answer
The percentage decrease in the partial pressure of \( O_2 \) is **58.33%**.