Question

In $100mL$ of water, a sample of hydrazine sulphate ( $N_2{H}_{6}S{O}_4$ ) 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:
$4F{e}^{3+}+N_2{H}_4\to N_2+4F{e}^{2+}+4H^{+}$

Answer 1

Answer: 13

${\text{Meqs of KMnO}}_4=\text{20}\times \frac{1}{\text{50}}\times 5=\text{2 Meqs}$
${\text{Meqs of KMnO}}_4\text{For}\text{ one Litre solution = 200 Meqs}$
$\cong \left[{\text{Meqs of Fe}}^{2+}\right]\cong \left[{\text{Meqs of N}}_2{\text{H}}_6{\text{SO}}_4\right]$
${\text{mM of N}}_2{\text{H}}_6{\text{SO}}_4=\frac{\text{200}}{\text{Z factor}}$
$\text{For }{\text{N}}_2{\text{H}}_6{\text{SO}}_4\to {\text{N}}_2+4{\text{H}}^{+}+{\text{H}}_2{\text{SO}}_4+4{\text{e}}^{-}$
4 moles of electrons transferred
per mole of ${\text{N}}_2{\text{H}}_6{\text{SO}}_4\text{in half reaction Z factor = 4}$
${\text{mM of N}}_2{\text{H}}_6{\text{SO}}_4=\frac{\text{200}}{\text{4}}=\text{50 mM}$
$=\text{50}\times 10^{-3}\times 130\text{g}/\text{mole}$
= 6.5g/Litre
So, For two liter $13g.$