Question

The number of moles of $B{r}_2$ 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

Answer 1

The correct option is A 5

When 2 moles of $KMn{O}_4$ reacts with excess of $KBr$,

$2KMn{O}_4+16H^{+}+10KBr\to 2M{n}^{2+}+12K^{+}+8H_{2}O+5B{r}_2$

5 moles of $Br$ is formed.

Hence, option A is correct.