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

Answer

A

Explanation

`Cr_(2)O_(7)^(2-) + 14H^(+) + 6e^(-) to 2Cr^(3+) + 7H_(2)O`
`(Sn^(2+) to Sn^(4+) + 2e^(-))xx3`
`Cr_(2)O_(7)^(2-) + 14H^(+) + 3Sn^(2+) to 3Sn^(4+) + 2Cr^(3+) + 7H_(2)O`
`implies 1mol Cr_(2)O_(7)^(2-) equiv 3mol of Sn^(2+)`
It is clear from this equation that 3 moles of `Sn^(2+)` reduce one mole of `Cr_(2)O_(7)^(2-),` hence 1 mol. of `Sn^(2+)` will reduce `1/3` moles of `Cr_(2)O_(7)^(2-)`.