0.75 mole of solid `A_(4)` and 2 mole of gaseous `O_(2)` are heated to react completely in a sealed bottle to produce gaseous compound `A_(3)O_(n)`. After the compound is formed, the vessel is brought to initial temperature, the pressure is found to half of initial pressure. The value of n is ________.

Explanation

To solve the problem, we need to analyze the reaction between solid A4 and gaseous O2 to produce the gaseous compound A3On. Let's break it down step by step.

### Step 1: Write the balanced reaction
We start with the reactants:
- Solid A4: 0.75 moles
- Gaseous O2: 2 moles

The reaction can be represented as:
\[ A_4 + O_2 \rightarrow A_3O_n \]

### Step 2: Determine the limiting reagent
To find the limiting reagent, we need to know how many moles of A3On can be produced from the available reactants.

From the stoichiometry of the reaction:
- 1 mole of A4 produces 1 mole of A3On.
- Each mole of O2 provides 2 moles of O atoms.

### Step 3: Calculate the moles of O atoms available
From 2 moles of O2:
\[ \text{Moles of O atoms} = 2 \text{ moles O}_2 \times 2 = 4 \text{ moles O} \]

### Step 4: Determine the moles of A3On produced
Since 0.75 moles of A4 can react with the available O atoms, we can find the moles of A3On produced:
- Each mole of A4 produces 1 mole of A3On, so:
\[ \text{Moles of A3On} = 0.75 \text{ moles of A4} \]

### Step 5: Calculate the total moles of gas after the reaction
Initially, we had:
- 2 moles of O2 (gaseous)
- 0 moles of A3On (gaseous)

After the reaction:
- Moles of A3On produced = 0.75 moles

Total moles of gas after the reaction:
\[ \text{Total moles of gas} = \text{Moles of A3On} = 0.75 \]

### Step 6: Analyze the pressure change
According to the problem, after the reaction, the pressure is found to be half of the initial pressure.

Using the ideal gas law, we know that pressure is directly proportional to the number of moles of gas at constant temperature and volume:
\[ \frac{P_1}{P_2} = \frac{n_1}{n_2} \]

Where:
- \( P_1 \) is the initial pressure,
- \( P_2 \) is the final pressure,
- \( n_1 \) is the initial number of moles of gas,
- \( n_2 \) is the final number of moles of gas.

Given that the final pressure is half the initial pressure:
\[ P_2 = \frac{P_1}{2} \]

This implies:
\[ n_2 = \frac{n_1}{2} \]

### Step 7: Calculate the initial number of moles of gas
Initially:
- Moles of gas (O2) = 2 moles

Thus:
\[ n_1 = 2 \text{ moles} \]

### Step 8: Find the final number of moles of gas
Since \( P_2 = \frac{P_1}{2} \):
\[ n_2 = \frac{2}{2} = 1 \text{ mole} \]

### Step 9: Set up the equation for O atoms
Using the principle of atom conservation:
- Total O atoms before = Total O atoms after

Before the reaction:
- O atoms from O2 = 4 (from 2 moles of O2)

After the reaction:
- O atoms in A3On = n (where n is the number of O atoms in the compound)

Thus:
\[ 4 = n \]

### Step 10: Conclusion
The value of n is 4. Therefore, the compound formed is \( A_3O_4 \).

### Final Answer:
The value of n is **4**.

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