Question

how to find the equivalent weight of K4[Fe(CN)6] in the reaction
K4[Fe(CN)6]--->(K^+ ) + (Fe^+3) +CO2+(NO3^-1).If the molar mass of K4[Fe(CN)6] is M

Answer 1

Answer:

first you have to find the valence factor for the Fe in K4[Fe(CN)6]

oxidation state of Fe is= +2

valence factor=3-2=1

Equivalent weight=M/1=M

Answer 2

The equivalent weight of K4[Fe(CiN)6]is M/4

We know that equivalent weight of a compound is equal to molecular mass divided by n factor.

nfactor of compound is equal to the valency of cation of anion of the compounds.

In the above compound the cation is potassium and since formula is formed by criss cross hence as CN is 4 so valency of cation that is potassium is 4.

Anion for reaction is [Fe(CN)6]ehose valency is 4- hence valency is 4.

Since valency of both cation and anion is 4 so

equivalent weight =molecular weight/valency.

Molecular weight of compound is M so

Equivalent weight=M/4