How many moles of `NO_(2)` are produced when 1 mole of `B` reacts with `HNO_(3)` ?

Explanation

To determine how many moles of \( NO_2 \) are produced when 1 mole of boron (B) reacts with \( HNO_3 \), we can follow these steps:

### Step 1: Write the balanced chemical equation
When boron reacts with nitric acid (\( HNO_3 \)), it forms boric acid (\( H_3BO_3 \)) and nitrogen dioxide (\( NO_2 \)). The balanced equation for this reaction is:

\[ B + 3 HNO_3 \rightarrow H_3BO_3 + 3 NO_2 \]

### Step 2: Analyze the coefficients
From the balanced equation, we can see that:
- 1 mole of boron produces 3 moles of \( NO_2 \).

### Step 3: Calculate the moles of \( NO_2 \)
Since we have 1 mole of boron reacting, we can directly use the stoichiometry from the balanced equation:

\[ \text{Moles of } NO_2 = 3 \times \text{Moles of } B \]

Given that we have 1 mole of boron:

\[ \text{Moles of } NO_2 = 3 \times 1 = 3 \text{ moles} \]

### Conclusion
When 1 mole of boron reacts with \( HNO_3 \), it produces **3 moles of \( NO_2 \)**.

---