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
B
3
C
2
D
4

Solution

## The correct option is A 5When 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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