Number of moles of `H_(2)` formed in the hydrolysis of a mole of borazole is :

Answer

To determine the number of moles of \( H_2 \) formed in the hydrolysis of one mole of borazole (\( B_3N_3H_6 \)), we can follow these steps:

### Step 1: Write the chemical formula of borazole
Borazole has the chemical formula \( B_3N_3H_6 \).

### Step 2: Understand the hydrolysis reaction
During hydrolysis, borazole reacts with water to form boric acid (\( H_3BO_3 \)), ammonia (\( NH_3 \)), and hydrogen gas (\( H_2 \)). The reaction can be represented as:
\[ B_3N_3H_6 + 6H_2O \rightarrow 3H_3BO_3 + 3NH_3 + 3H_2 \]

### Step 3: Analyze the stoichiometry of the reaction
From the balanced equation, we can see that for every mole of borazole that reacts, 3 moles of \( H_2 \) are produced.

### Step 4: Calculate the number of moles of \( H_2 \) produced
Since we start with 1 mole of borazole, the number of moles of \( H_2 \) produced can be calculated as follows:
\[ \text{Moles of } H_2 = 1 \text{ mole of } B_3N_3H_6 \times \frac{3 \text{ moles of } H_2}{1 \text{ mole of } B_3N_3H_6} = 3 \text{ moles of } H_2 \]

### Conclusion
Thus, the number of moles of \( H_2 \) formed in the hydrolysis of one mole of borazole is **3 moles**.

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