Question

For,$KMn{O_4}$ reacts oxalic acid according to the equations :-
$2Mn{O_4}^ - + 5{C_2}{O_4}^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$
20 ml of $0.1M$ $KMn{O_4}$ Will you react ?
(A). 120 ml of 0.25 M oxalic acid .
(B). 150 ml of 0.1 M oxalic acid .
(C). 50 ml of 0.1 M oxalic acid .
(D). 150 ml of 0.2 M oxalic acid .

Answer 1

Hint:
According to the equation ,
$2Mn{O_4}^ - + 5{C_2}{O_4}^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$
2 moles of $Hn{O_4}^ - $ are reacting with 5 moles of ${C_2}{O_4}^{2 - }$ . We will apply the same approach to solve questions .

Complete step by step answer:
$KMn{O_4}$ Potassium permanganate is a strong oxidizing agent and in the pressure of sulfuric acid it acts as a powerful oxidizing agent . In acidic solution .
$Mn{O_4}^ - + 8{H^ + } + 5{e^ - } \to H{n^{2 + }} + 4{H_2}O$
Generally , the solutions containing $Mn{O_4}^ - $ ions are purple in colour and the solution having $M{n^{2 + }}$ ions are colourless and hence it can be concluded that the permanganate solution is decolourized when added to the solution of a reducing agent .
Potassium permanganate is thus known as a self indicator .
The $KMn{O_4}$ with oxalic acid is used in the titration process . Now , according to the equation given in question that is ,
$2Mn{O_4}^ - + 5{C_2}{O_4}^ - + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$
2 moles of $Mn{O_4}^ - $ is reacting with 5 moles of ${C_2}{O_4}^ - $ .
At NTP,
$2 \times 22.4L$ of $Mn{O_4}^ - $ is reacting with $5 \times 22.4L$ of ${C_2}{O_4}^ - $ .
Therefore , 20 ml of $0.1mKMn{O_4}$ will react with oxalic acid
$ = \dfrac{{5 \times 22.4 \times 20}}{{2 \times 22.4}}$
$ = 50$ ml
Hence , the correct option is C.
20 ml of $0.1KMn{O_4}$ will react with 50 ml of 0.1 m oxalic acid .

Note:The reaction of potassium permanganate (${\text{KMn}}{{\text{O}}_{\text{4}}}$) with oxalic acid is a redox reaction in which the oxalic acid is oxidised to carbon dioxide by potassium permanganate and potassium permanganate itself got reduce to ${\text{Mn}}{{\text{O}}_{\text{4}}}$.