How many millitres of `0.05M K_(4)[Fe(CN)_(6)]` solution is required for titration of `60ml` of `0.01m ZnSO_(4)` solutioin, when the product of reactgion is `K_(2)Zn_(3)[Fe(CN)_(6)]_(2)`?

Answer

8

Explanation

`2K_(4)[Fe(CN)_(6)]+3ZnSO_(4) rarr K_(2)Zn_(3)[Fe(CN)_(6)]_(2) + 3K_(2)SO_(4), 2 mol K_(4)[Fe(CN)_(6)] -= 3m mol ZnSO_(4)`
`0.05 xx V_(mL)K_(4)[Fe(CN)_(6)] -= (3)/(2) xx 0.05 xx V_(mL) ZnSO_(4) = 60 xx 0.01 rArr V_(mL) = 8 mL`