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Question

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:


A
5
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B
3
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C
2
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D
4
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Solution

The correct option is A 5
When 2 moles of $$KMnO_4$$ reacts with excess of $$KBr$$,

$$\displaystyle 2KMnO_4 + 16 H^++10KBr \rightarrow 2Mn^{2+} +  12 K^+ +8H_2O+5Br_2$$

5 moles of $$Br$$ is formed.

Hence, option A is correct.

Chemistry

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