The number of moles of `Br_(2)` produced when two moles of potassium permanganate are treated with excess potassium bromide in aqueous acid medium is-

Options

A. 1
B. 3
C. 2
D. 4

Explanation

To solve the problem of determining the number of moles of \( \text{Br}_2 \) produced when two moles of potassium permanganate (\( \text{KMnO}_4 \)) are treated with excess potassium bromide (\( \text{KBr} \)) in an aqueous acid medium, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between potassium permanganate and potassium bromide in an acidic medium can be represented as:
\[ 2 \text{KMnO}_4 + 10 \text{KBr} + 8 \text{H}_2\text{SO}_4 \rightarrow 2 \text{MnSO}_4 + 5 \text{Br}_2 + 6 \text{K}_2\text{SO}_4 + 8 \text{H}_2\text{O} \]

### Step 2: Identify the stoichiometry of the reaction
From the balanced equation, we can observe the stoichiometric coefficients:
- 2 moles of \( \text{KMnO}_4 \) produce 5 moles of \( \text{Br}_2 \).

### Step 3: Calculate the moles of \( \text{Br}_2 \) produced
Given that we start with 2 moles of \( \text{KMnO}_4 \):
\[ \text{Moles of } \text{Br}_2 = \left( \frac{5 \text{ moles of } \text{Br}_2}{2 \text{ moles of } \text{KMnO}_4} \right) \times 2 \text{ moles of } \text{KMnO}_4 = 5 \text{ moles of } \text{Br}_2 \]

### Conclusion
Thus, the number of moles of \( \text{Br}_2 \) produced is **5 moles**.

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