1.

n_(1)Cr_(2)O_(7)^(-2) + n_(2)FeC_(2)O_(4) underset("Medium")overset("Acidic")rarr n_(3)Cr^(+3) + n_(4)Fe^(+3) + n_(5)CO_(2) + n_(6)H_(2)O here n_(1), n_(2), n_(3), n_(4), n_(5) and n_(6) are simplest non fractional coefficients of the balanced redox reaction. Then value of [n_(1) + n_(2) + n_(3) + n_(4) + n_(5) + n_(6)] is :

Answer»


SOLUTION :`14H^(+) + Cr_(2)O_(7)^(-2) + 2FeC_(2)O_(4) RARR 2Cr^(+3) + 2Fe^(+3) + 4CO_(2) + 7H_(2)O`
`therefore n_(1)= 1, n_(2)=2, n_(3)= 2, n_(4)= 2, n_(5)= 4, n_(6)= 7`
And `(n_(1) + n_(2) + n_(3) + n_(4) + n_(5) + n_(6))= 18`


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