Question

In $100mL$ of water, a sample of hydrazine sulphate ( $N_{2}H_{6}SO_{4}$ ) was dissolved, $10mL$ of this solution was reacted with excess of ferric chloride solution and warmed to complete the reaction. Ferrous ion formed was estimated and it required $20mLofM/50$ potassium permanganate solution. Find out the amount of hydrazine sulphate in two litre of the solution. The reaction is given below:
$4Fe^{3 +}+N_{2}H_{4} \rightarrow N_{2}+4Fe^{2 +}+4H^{+}$

Answer 1

$\text{Meqs of KMnO}_{4} = \text{20} \times \frac{1}{\text{50}} \times 5 = \text{2 Meqs}$
$\text{Meqs of KMnO}_{4} \text{For} \text{ one Litre solution = 200 Meqs}$
$\cong \left[\text{Meqs of Fe}^{2 +}\right] \cong \left[\text{Meqs of N}_{2} \text{H}_{6} \text{SO}_{4}\right]$
$\text{mM of N}_{2} \text{H}_{6} \text{SO}_{4} = \frac{\text{200}}{\text{Z factor}}$
$\text{For } \text{N}_{2} \text{H}_{6} \text{SO}_{4} \rightarrow \text{N}_{2} + 4 \text{H}^{+} + \text{H}_{2} \text{SO}_{4} + 4 \text{e}^{-}$
4 moles of electrons transferred
per mole of $\text{N}_{2} \text{H}_{6} \text{SO}_{4} \text{in half reaction Z factor = 4}$
$\text{mM of N}_{2} \text{H}_{6} \text{SO}_{4} = \frac{\text{200}}{\text{4}} = \text{50 mM}$
$= \text{50} \times 1 0^{- 3} \times 1 3 0 \text{g} / \text{mole}$
= 6.5g/Litre
So, For two liter $13g.$