One mole of acidified `K_(2)Cr_(2)O_(7)` on reaction with excess of KCl will liberate….., moles of `I_(2)`.

Answer

3

Explanation

`Cr_(2)O_(7)^(2-)+14H^(+)+6I^(-)rarr2Cr^(3^(+))+7H_(2)O+3I_(2)`
3 moles of `I_(2)` per mole of `K_(2)Cr_(2)O_(7)` are produced.