Question

A sample of hydrazine sulphate $\left( {{{\text{N}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{S}}{{\text{O}}_{\text{4}}}} \right)$ was dissolved in $100{\text{ mL}}$ of water and $10{\text{ mL}}$ 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 $20{\text{ mL}}$ of $\dfrac{M}{{50}}$ potassium permanganate solution. Estimate the amount of hydrazine sulphate in $1{\text{ L}}$ of the solution. Reactions are given below:
${\text{4F}}{{\text{e}}^{3 + }} + {{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}} \to {{\text{N}}_{\text{2}}} + {\text{4F}}{{\text{e}}^{2 + }} + {\text{4}}{{\text{H}}^ + }$
${\text{MnO}}_4^ - + {\text{5F}}{{\text{e}}^{2 + }} + {\text{8}}{{\text{H}}^ + } \to {\text{M}}{{\text{n}}^{2 + }} + {\text{5F}}{{\text{e}}^{3 + }} + {\text{4}}{{\text{H}}_{\text{2}}}{\text{O}}$
(A) $1.91{\text{ g}}$
(B) $3.82{\text{ g}}$
(C) $2.71{\text{ g}}$
(D) $6.50{\text{ g}}$

Answer 1

First calculate the normality of $20{\text{ mL}}$ of $\dfrac{M}{{50}}$ potassium permanganate solution. For this, carefully calculate the valency factor for potassium permanganate. Then from the reaction stoichiometry, calculate the equivalent mass of hydrazine sulphate. Then calculate the amount of hydrazine sulphate.

Complete step-by-step solution: We are given the reactions as follows:
${\text{4F}}{{\text{e}}^{3 + }} + {{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}} \to {{\text{N}}_{\text{2}}} + {\text{4F}}{{\text{e}}^{2 + }} + {\text{4}}{{\text{H}}^ + }$
${\text{MnO}}_4^ - + {\text{5F}}{{\text{e}}^{2 + }} + {\text{8}}{{\text{H}}^ + } \to {\text{M}}{{\text{n}}^{2 + }} + {\text{5F}}{{\text{e}}^{3 + }} + {\text{4}}{{\text{H}}_{\text{2}}}{\text{O}}$
The relation between normality and molarity is as follows:
${\text{Normality}} = n \times {\text{Molarity}}$
Where $n$ is the valency factor i.e. the change in oxidation state of an atom in a compound per molecule.
From the reaction, we can see that ${\text{MnO}}_4^ -$ changes to ${\text{M}}{{\text{n}}^{2 + }}$. Thus, the oxidation state changes from $+ 7$ to $+ 2$. Thus, the valency factor for ${\text{KMn}}{{\text{O}}_{\text{4}}}$ is 5.
Thus, the normality of $20{\text{ mL}}$ of $\dfrac{M}{{50}}$ potassium permanganate solution is,
${\text{Normality}} = 5 \times {\text{20 mL}}\dfrac{M}{{50}}{\text{KMn}}{{\text{O}}_4}$
${\text{Normality}} = {\text{20 mL}}\dfrac{N}{{10}}{\text{KMn}}{{\text{O}}_4}$
From the given reactions, we can see that,
${\text{20 mL}}\dfrac{N}{{10}}{\text{KMn}}{{\text{O}}_4} \equiv {\text{20 mL}}\dfrac{N}{{10}}{\text{F}}{{\text{e}}^{2 + }} \equiv {\text{20 mL}}\dfrac{N}{{10}}{\text{FeC}}{{\text{l}}_3} \equiv {\text{20 mL}}\dfrac{N}{{10}}{{\text{N}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{S}}{{\text{O}}_{\text{4}}}$
From the reaction, we can see that ${{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}$ changes to ${{\text{N}}_{\text{2}}}$. Thus, the oxidation state changes from $+ 4$ to 0. Thus, the valency factor for ${{\text{N}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is 4.
Thus, the equivalent mass of ${{\text{N}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is,
${\text{Equivalent mass}} = \dfrac{{{\text{Molar mass}}}}{4}$
${\text{Equivalent mass}} = \dfrac{{{\text{130}}}}{4} = 32.5$
Thus, the amount of hydrazine sulphate i.e. ${{\text{N}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{S}}{{\text{O}}_{\text{4}}}$ in $10{\text{ mL}}$ solution is,
Amount of hydrazine sulphate $= \dfrac{1}{{10}} \times \dfrac{{32.5}}{{1000}} \times 20 = 0.065{\text{ g}}$
Thus, the amount of hydrazine sulphate i.e. ${{\text{N}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{S}}{{\text{O}}_{\text{4}}}$ in one litre solution is,
Amount of hydrazine sulphate $= \dfrac{{0.065{\text{ g}}}}{{10{\text{ mL}}}} \times 1000{\text{ mL}} = 6.50{\text{ g}}$
Thus, the amount of hydrazine sulphate in $1{\text{ L}}$ of the solution is $6.50{\text{ g}}$.

Thus, the correct option is (D) $6.50{\text{ g}}$.

Note: Remember that the valency factor is the change in oxidation state of an atom in a compound per molecule. When ${\text{MnO}}_4^ -$ changes to ${\text{M}}{{\text{n}}^{2 + }}$, the oxidation state changes from $+ 7$ to $+ 2$. Thus, the valency factor for ${\text{KMn}}{{\text{O}}_{\text{4}}}$ is 5. When ${{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}$ changes to ${{\text{N}}_{\text{2}}}$ the oxidation state changes from $+ 4$ to 0. Thus, the valency factor for ${{\text{N}}_{\text{2}}}{{\text{H}}_{\text{6}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is 4.