The number of moles of `K_(2)Cr_(2)O_(7)` reduced by `1 mol` of `Sn^(2+)` ions is

Options

A. `1//3`
B. 3
C. `1//6`
D. 6

Explanation

`(6e^(-)+Cr_(2)O_(7)^(2-) to 2Cr^(3+) Sn^(2+) to Sn^(4+) +2e^(-)`
Equivalent of `Cr_(2)O_(7)^(2-)="Equivalent of "Sn^(+2)`
(n=6) (n=2) `rArr 1Eq=1 Eq`
`1/6" mol" =1/2" mol"`
`1/3" mol" Cr_(2)O_(7)^(2-)="1 mol of Sn"^(2+)`