1.

How many moles of K_(2)Cr_(2)O_(7)can be reduced by 1 mole of Sn^(2+)?

Answer»


SOLUTION :`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-)`.


Discussion

No Comment Found

Related InterviewSolutions