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

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 forF{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.