Potassium iodide reacts with acidified `K_(2)Cr_(2)O_(7)`. How many moles of KI are required for one mole of `K_(2)Cr_(2)O_(7)` ?

Answer

`(6)`

Explanation

`[Cr_(2)O_(7)^(2-)+14H^(o+)+6e^(Θ)rarr 2Cr^(3+)+7H_(2)O]`
`2I^(Θ)rarr I_(2)+2e^(Θ)]xx3`
`ulbar(Cr_(2)O_(7)^(2-)+14H^(o+)+61^(Θ)rarr 2Cr^(3+)+7H_(2)O+3I_(2)`
1 mol of `K_(2)Cr_(2)O_(7)` reacts with 6 moles of KI.