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One mole of acidified '(K)(2) (Cr)(2) (O...

One mole of acidified '(K)_(2) (Cr)_(2) (O)_(7)' on reaction with excess of '(Kl)' will liberate moles of '(I)'

A

6

B

1

C

7

D

3

Text Solution

Verified by Experts

The correct Answer is:
D
`Cr_(2)O_(7)""^(2-)+14H^(+)+6e^(-) rarr 2Cr^(3+)+7H_(2)O (2I^(-) rarr I_(2)+2e^(-)) times 3`
`Cr_(2)O_(7)""^(2-)+14H^(+)+6I^(-) rarr 2Cr^(3+)+7H_(2)O+3I_(2)`
Hence, number of moles of `I_(2)` produced = 3
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