Question

How many coulomb of electricity would be required to reduce the iron in $36.0g$ of potassium hexacyano ferrate (III), $K_{3}Fe(CN{)}_6$, to metallic iron?

Answer 1

$F{e}^{3+}$ + $3e^{-}$ $\to$ $Fe$

that means, 3F of current is require to deposit 1 mole of $Fe$

number of moles of $K_{3}Fe(CN{)}_6$ = $\frac{givenweight}{moleculrweight}$

no of moles = $\frac{36}{329}$ $=0.11$ moles

for 1 mole it require 3 F ($3\ast 96500$)

for 0.11 mole current require = $3\ast 96500\ast 0.11$

so, number of coulombs require = 32134.5