1.

What is the equivalent weight of KH(IO_(3))_(2) as an oxidant in presence of 4.0 (N) HCl when Icl becomes the reduced form ? (K=39.0, I=127.0).

Answer»

Solution :In `KH(IO_(3))_(2),IO_(3)` is present as `IO_(3)^(-)`. OXIDATION state of I will be `x-6=-1 or x=+5`
`overset(+5)(KH(IO_(3))_(2))RARR overset(+1)(2ICl)`
Decrease in oxidation state `=10-2=8`
`therefore"Equivalent wt. of "KH(IO_(3))_(2)=("MOL. wt.")/(8)=(39+1+2(127+48))/(8)=48.45`


Discussion

No Comment Found

Related InterviewSolutions