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# The reaction of potassium permanganate with acidified iron $\left[ {{\rm{II}}} \right]$sulphate is given below: ${\rm{2KMn}}{{\rm{O}}_{\rm{4}}}{\rm{ + 10}}\,{\rm{FeS}}{{\rm{O}}_{\rm{4}}}{\rm{ + 8}}\,{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} \to {{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{ + 2}}\,{\rm{MnS}}{{\rm{O}}_{\rm{4}}}{\rm{ + 5}}\,{\rm{F}}{{\rm{e}}_{\rm{2}}}{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)_{\rm{3}}}{\rm{ + 8}}\,{{\rm{H}}_{\rm{2}}}{\rm{O}}$If ${\rm{15}}{\rm{.8}}\,{\rm{g}}$of potassium permanganate was used in the reaction, calculate the mass of iron $\left[ {{\rm{II}}} \right]$sulphate used in the above reaction.Molecular weight of ${\rm{K}}\,{\rm{ = }}\,{\rm{39}}$,${\rm{Mn}}\,{\rm{ = }}\,{\rm{55}}$,${\rm{Fe}}\,{\rm{ = }}\,{\rm{56}}$, ${\rm{S}}\,{\rm{ = }}\,{\rm{32}}$, ${\rm{O}}\,{\rm{ = }}\,{\rm{16}}\,\dfrac{{\rm{g}}}{{{\rm{mole}}}}$(A) ${\rm{76}}\,{\rm{g}}$(B) $152\,{\rm{g}}$(C) $38\,{\rm{g}}$(D) ${\rm{46}}\,{\rm{g}}$

Last updated date: 13th Jun 2024
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Hint:As we know that the above given reaction is a redox reaction in which oxidation of iron and reduction of manganese occurs simultaneously. the mass of iron $\left[ {{\rm{II}}} \right]$sulphate can be estimated by calculating that one mole of potassium permanganate reacts with how many moles of iron $\left[ {{\rm{II}}} \right]$sulphate.

Complete step-by-step solution:we know that a redox reaction is a reaction in which electron transfer occurs between different species causing oxidation and reduction simultaneously. The oxidation number of a specific atom, molecule or ion changes by the loss or gain of electrons.
So, this given reaction is a redox reaction because in this reaction manganese is reduced and the mass of iron $\left[ {{\rm{II}}} \right]$sulphate is oxidized. The reaction is represented as-
${\rm{2KMn}}{{\rm{O}}_{\rm{4}}}{\rm{ + 10}}\,{\rm{FeS}}{{\rm{O}}_{\rm{4}}}{\rm{ + 8}}\,{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} \to {{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{ + 2}}\,{\rm{MnS}}{{\rm{O}}_{\rm{4}}}{\rm{ + 5}}\,{\rm{F}}{{\rm{e}}_{\rm{2}}}{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)_{\rm{3}}}{\rm{ + 8}}\,{{\rm{H}}_{\rm{2}}}{\rm{O}}$
In this given reaction, we can see that
$2$mole of ${\rm{KMn}}{{\rm{O}}_4}$ is reacting with $10$mole of iron $\left[ {{\rm{II}}} \right]$sulphate so,
${\rm{1}}\,{\rm{mole}}\,$ of ${\rm{KMn}}{{\rm{O}}_4}$will react with ${\rm{ = }}\,5\,{\rm{mole}}\,$of iron $\left[ {{\rm{II}}} \right]$sulphate
Now we are given ${\rm{15}}{\rm{.8}}\,{\rm{g}}$of ${\rm{KMn}}{{\rm{O}}_4}$, from this value we can calculate number of moles of ${\rm{KMn}}{{\rm{O}}_4}$as
${\rm{number}}\,{\rm{of}}\,{\rm{moles}} = \dfrac{{{\rm{given}}\,{\rm{weight}}\,}}{{{\rm{molecular}}\,{\rm{weight}}}}$
Where molecular weight of ${\rm{KMn}}{{\rm{O}}_{\rm{4}}} = {\rm{158}}\,\,\dfrac{{\rm{g}}}{{{\rm{mole}}}}$
By Putting the above values in above equation, we get as
$\begin{array}{c} {\rm{number}}\,{\rm{of}}\,{\rm{moles}} = \dfrac{{{\rm{15}}{\rm{.8}}\,{\rm{g}}\,}}{{{\rm{158}}\,{{\rm{g}} \mathord{\left/ {{{\rm{g}} {{\rm{mole}}}}} \right. } {{\rm{mole}}}}}}\\ = {\rm{0}}{\rm{.1}}\,{\rm{mole}} \end{array}$
Therefore, ${\rm{0}}{\rm{.1}}\,{\rm{mole}}\,$ of ${\rm{KMn}}{{\rm{O}}_4}$will react with $= {\rm{(5 \times }}\,{\rm{0}}{\rm{.1}}\,{\rm{)}}\,{\rm{mole}} = {\rm{0}}{\rm{.5}}\,{\rm{mole}}$of iron $\left[ {{\rm{II}}} \right]$sulphate
The options are given in gram so we will calculate mass of iron $\left[ {{\rm{II}}} \right]$sulphate as
$\begin{array}{c} {\rm{mass}}\,{\rm{of}}\,{\rm{Fe(S}}{{\rm{O}}_{\rm{4}}}{\rm{)}} = {\rm{0}}{\rm{.5}}\,{\rm{mole}} \times \,{\rm{152}}\,{{\rm{g}} \mathord{\left/ {{{\rm{g}} {{\rm{mole}}}}} \right. } {{\rm{mole}}}}\\ = {\rm{76}}\,{\rm{g}} \end{array}$

Therefore, the correct option is option (A).

Note:The above method is the easy method to calculate the mass of reactant or product. Only we have to know the balanced reaction. Some questions do not contain any reaction so we have to use the law of conservation of mass to balance the reacting species and products.