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 will follow these steps:

### Step 1: Identify the reaction
We start with 0.75 moles of solid \( A_4 \) and 2 moles of gaseous \( O_2 \). The reaction can be represented as:
\[ x A_4 + y O_2 \rightarrow z A_3O_n \]

### Step 2: Determine the limiting reactant
Since we have 0.75 moles of \( A_4 \) and 2 moles of \( O_2 \), we need to find out how many moles of \( O_2 \) are required to react completely with \( A_4 \).

Assuming the reaction stoichiometry is \( A_4 + O_2 \rightarrow A_3O_n \), we can determine the number of moles of \( O_2 \) required for 0.75 moles of \( A_4 \).

### Step 3: Calculate moles of products formed
From the reaction stoichiometry, if 1 mole of \( A_4 \) reacts with \( y \) moles of \( O_2 \), then 0.75 moles of \( A_4 \) would react with \( 0.75y \) moles of \( O_2 \).

Given that we have 2 moles of \( O_2 \), we can conclude that \( A_4 \) is the limiting reactant since 0.75 moles of \( A_4 \) will require less than 2 moles of \( O_2 \).

### Step 4: Determine the number of moles of \( A_3O_n \) produced
Let’s assume that the reaction produces \( z \) moles of \( A_3O_n \). Since \( A_4 \) is the limiting reactant, the number of moles of \( A_3O_n \) produced will be equal to the number of moles of \( A_4 \) reacted, which is 0.75 moles.

### Step 5: Analyze the pressure change
Initially, the pressure \( P_1 \) is due to the 2 moles of \( O_2 \):
\[ P_1 \propto 2 \text{ moles of } O_2 \]

After the reaction, the pressure \( P_2 \) is due to the moles of \( A_3O_n \) formed. We know that after cooling, the pressure is half of the initial pressure:
\[ P_2 = \frac{P_1}{2} \]

This implies:
\[ \frac{P_1}{P_2} = 2 \implies \frac{2}{z} = 2 \implies z = 1 \]

### Step 6: Determine the value of \( n \)
From the stoichiometry of the reaction, we have:
- From 0.75 moles of \( A_4 \), we get \( 0.75 \times 4 = 3 \) moles of \( A \).
- From 2 moles of \( O_2 \), we get \( 2 \) moles of \( O \).

Thus, the total number of moles of \( A \) and \( O \) in the product \( A_3O_n \) is:
\[ 3A + 2O \implies n = 4 \]

### Final Answer
The value of \( n \) is **4**.

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