Question

$KMn{O_4}$ reacts with oxalic acid according to the equation
$2Mn{O_4}^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$
Here, $20ml$ of $0.1M$ $KMn{O_4}$ is equivalent to:
(This question has multiple correct options)
A. $120ml$ of $0.25M$ ${H_2}{C_2}{O_4}$
B. $50ml$ of $0.10M$ ${H_2}{C_2}{O_4}$
C. $25ml$ of $0.20M$ ${H_2}{C_2}{O_4}$
D. $50ml$ of $0.20M$ ${H_2}{C_2}{O_4}$

Answer 1

The following given reaction is the redox reaction. In order to balance such redox reactions stoichiometry calculation is done. Stoichiometry is done to determine the relation between reactants and products quantitatively. Simply it is the measure of elements in the reaction.

Complete step by step answer:
According to the question $KMn{O_4}$ reacts with oxalic acid (followed by the given equation):
$2Mn{O_4}^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$
By stoichiometry, $2moles$ of $KMn{O_4}$ is equivalent to $5moles$ of ${H_2}{C_2}{O_4}$ . As $20ml$ of $0.1M$ $KMn{O_4}$ contains $0.1 \times 0.020moles$ of $KMn{O_4}$
$\Rightarrow 0.002moles$
So, $0.002moles$ of $KMn{O_4}$ will be equivalent to $0.005moles$ of ${H_2}{C_2}{O_4}$
Now, in the option A:
$120ml$ of $0.25M$ ${H_2}{C_2}{O_4}$
Number of moles of ${H_2}{C_2}{O_4}$ is equal to $0.25 \times 0.12moles$
$\Rightarrow 0.03moles$
It is not equivalent to $0.002moles$ of $KMn{O_4}$ so this is not the correct option.
Now, in the option B:
$50ml$ of $0.10M$ ${H_2}{C_2}{O_4}$
Number of moles of ${H_2}{C_2}{O_4}$ is equal to $0.10 \times 0.50moles$
$\Rightarrow 0.005moles$
It is equivalent to $0.002moles$ of $KMn{O_4}$ so this is the correct option.
Now, in the option C:
$25ml$ of $0.20M$ ${H_2}{C_2}{O_4}$
Number of moles of ${H_2}{C_2}{O_4}$ is equal to $0.20 \times 0.25moles$
$\Rightarrow 0.005moles$
It is equivalent to $0.002moles$ of $KMn{O_4}$ so this is also a correct option.
Now, in the option D:
$50ml$ of $0.20M$ ${H_2}{C_2}{O_4}$
Number of moles of ${H_2}{C_2}{O_4}$ is equal to $0.20 \times 0.50moles$
$\Rightarrow 0.01moles$
It is equivalent to $0.002moles$ of $KMn{O_4}$ so this is not the correct option.

So, option B and option C are correct.
Note:
Redox reactions are such types of reactions where oxidation and reduction both take place simultaneously. In redox reactions oxidation states of atoms get changed. In this reaction one species undergoes oxidation and on the other hand other species undergoes reduction. The oxidation half is identified by using the oxidation state such as increase in oxidation state or loss of electron is oxidation and the reduction half is identified where oxidation state decreases and gain of electrons takes place.