Question

one mole each \[Fe{C_2}{O_4}\,\], \[F{e_2}{({C_2}{O_4})_3}\],\[\,\,FeS{O_4}\] ,\[F{e_2}{(S{O_4})_3}\], will react with how many moles of acidified $KMn{O_4}$​?
(A) $1$
(B) $2$
(C) $3$
(D) $5$

Answer 1

Hint:To find the total number of moles of $KMn{O_4}$ we have to calculate the n factor value of $KMn{O_4}$ by comparing the number of gram equivalent of Oxidising agent and the number of gram equivalent of reducing agent.


Complete solution:
 Now we will compare the number of gram equivalent of oxidizing agent which is $KMn{O_4}$ and the number of gram equivalent of reducing agents which are \[Fe{C_2}{O_4}\,\], \[F{e_2}{({C_2}{O_4})_3}\],\[\,\,FeS{O_4}\] ,\[F{e_2}{(S{O_4})_3}\]. Number of gram equivalent is given as:
\[Mole{s_{O.A.}}\,\, \times \,\,n{{ }}facto{r_{O.A.}}\; = Mole{s_{R.A.}}{{ }} \times n{{ }}facto{r_{R.A.}}\]\[{M_{KMn{o_4}}} \times \,\,n.{f_{KMn{o_4}}} = ({M_{Fe{C_2}{O_4}\,}} \times \,\,n.{f_{Fe{C_2}{O_4}\,}})\, + ({M_{F{e_2}{{({C_2}{O_4})}_3}\,}} \times \,\,n.{f_{F{e_2}{{({C_2}{O_4})}_3}\,}})\, + ({M_{FeS{O_4}\,}} \times \,\,n.{f_{FeS{O_4}\,}}) + ({M_{\,F{e_2}{{(S{O_4})}_3}\,}} \times \,\,n.{f_{F{e_2}{{(S{O_4})}_3}\,}})\]Now we will calculate the n factor of $KMn{O_4}$. In acidic medium the reaction of $KMn{O_4}$ reduces from $MnO_4^ - $ to $M{n^{2 + }}$ the reaction for the same is:
$MnO_4^ - \, + \,\,8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$
n Factor for $KMn{O_4}$$ = 5$, Now we will calculate the n factor of \[Fe{C_2}{O_4}\,\].
In \[Fe{C_2}{O_4}\,\]Iron is in $ + 2$ oxidation state, so it will get oxidised to achieve the highest $ + 3$Oxidation state. The reaction will be:
\[Fe{C_2}{O_4}\, + \,\,KMn{O_4}\, \to F{e^{3 + }} + \,\,C{O^{2 + }} + M{n^{2 + }}\]
In this $F{e^{2 + }} \to F{e^{3 + }} + 1{e^ - }$ and ${C_2}O_4^{2 - } \to 2C{O_2} + 2e$ so n Factor for \[Fe{C_2}{O_4}\,\]$ = 3$
Now we will calculate the n factor of \[F{e_2}{({C_2}{O_4})_3}\]. In \[F{e_2}{({C_2}{O_4})_3}\] Iron is in $ + 2$ oxidation state, so it will get oxidised to achieve the highest $ + 3$ Oxidation state. The reaction will be:
\[F{e_2}{({C_2}{O_4})_3}\, + \,\,KMn{O_4}\, \to F{e^{3 + }} + \,\,C{O^{2 + }} + M{n^{2 + }}\]
In this $F{e^{2 + }} \to F{e^{3 + }} + 1{e^ - }$and ${C_2}O_4^{2 - } \to 2C{O_2} + 2e$ and there are two iron atoms so n Factor for \[F{e_2}{({C_2}{O_4})_3}\]$ = 6$.
Now we will calculate the n factor of \[\,\,FeS{O_4}\]. In \[\,\,FeS{O_4}\]Iron is in $ + 2$ oxidation state, so it will get oxidised to achieve the highest $ + 3$Oxidation state. The reaction will be:
\[\,\,FeS{O_4}\, + \,\,KMn{O_4}\, \to F{e^{3 + }} + \,\,S{O_4}^{2 - } + M{n^{2 + }}\]
In this $F{e^{2 + }} \to F{e^{3 + }} + 1{e^ - }$and so n Factor for\[\,\,FeS{O_4}\]$ = 1$
\[F{e_2}{(S{O_4})_3}\] do not oxidise. So its n factor will be zero.
Now putting these values in the gram equivalent equation then we will get:

\[{M_{KMn{o_4}}} \times \,\,n.{f_{KMn{o_4}}} = ({M_{Fe{C_2}{O_4}\,}} \times \,\,n.{f_{Fe{C_2}{O_4}\,}})\, + ({M_{F{e_2}{{({C_2}{O_4})}_3}\,}} \times \,\,n.{f_{F{e_2}{{({C_2}{O_4})}_3}\,}})\, + ({M_{FeS{O_4}\,}} \times \,\,n.{f_{FeS{O_4}\,}}) + ({M_{\,F{e_2}{{(S{O_4})}_3}\,}} \times \,\,n.{f_{F{e_2}{{(S{O_4})}_3}\,}})\]
\[{M_{KMn{o_4}}} \times \,\,5 = (1 \times \,\,3)\, + (1 \times \,\,6)\, + (1 \times \,\,1) + (1 \times \,\,0)\]
\[{M_{KMn{o_4}}} = \dfrac{{10}}{5} = 2\]
Hence, the total number of $KMn{O_4}$ moles required is $2$.

Thus, the correct option is (B) .


Note:$KMn{O_4}$ or Potassium permanganate is an organic compound. It is used as an oxidizing agent. It is purple-black crystalline solid in colour. It reduces itself and oxidizes the other compounds. It is soluble in water and is also used for cleaning of wounds.