The number of moles of $B r_{2}$ produced when two moles of potassium permanganate is treated with excess potassium bromide in aqueous acid medium is:

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Solution

The correct option is A 5
In acid medium , $K M n O_{4}$ undergoes a change in its oxidation number from +7 to +2 , i.e it forms ${M n}^{2 +}$
so equation representing above reaction is $K M n O_{4}$ + KBr $\to$ ${M n}^{2 +}$ + ${B r}_{2}$

Change in oxidation number of with respect to central metal of $K M n O_{4}$ is 5 (change from +7 to +2 )

in KBr oxidation number of bromine is -1 and in ${B r}_{2}$ oxidation number of bromine is 0 , so change in oxidation number with respect to ${B r}_{2}$ is 1 but since there are two bromine atoms in ${B r}_{2}$ so change in oxidation number per atom with respect to ${B r}_{2}$ is 2

Applying Law of equivalence ; no. of equivalents of $K M n O_{4}$ = no. of equivalents of ${B r}_{2}$
No. of equivalents = moles x valency factor( n-factor)

n-factor= change in oxidation number per atom

hence, no. of equivalents of $K M n O_{4}$ = no. of equivalents of ${B r}_{2}$
moles $K M n O_{4}$ x n-factor = moles of ${B r}_{2}$ x n-factor
2 x 5 = moles of ${B r}_{2}$ x 2
hence moles of ${B r}_{2}$ = 5 .

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