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Oxidising Reagent and Reducing Reagent

In the reacti...

Question

In the reaction of $K M n O_{4}$ with $H_{2} C_{2} O_{4}$ , $20$ mL of $0.2 M K M n O_{4}$ is not equivalent to (in acidic medium):

A

120

mL of

0.25 M H_{2} C_{2} O_{4}

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B

100

mL of

0.1 M H_{2} C_{2} O_{4}

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C

500

mL of

0.2 M H_{2} C_{2} O_{4}

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D

10

mL of

1 M H_{2} C_{2} O_{4}

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Solution

The correct options are
A $120$ mL of $0.25 M H_{2} C_{2} O_{4}$
D $500$ mL of $0.2 M H_{2} C_{2} O_{4}$
As per following reaction ionic equation:
$2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$ . This shows for 2 mol of $M n O 4^{2 -}$ , 5 moles of $C_{2} O_{4}^{2 -}$ are required. so, As number of moles present in 20 mL of 0.2 M $K M n O_{4}$ = 20 * 0.2/1000 = $4 / 1000$ moles Hence as per equation no. of moles of $H_{2} C_{2} O_{4}$ required = 5 * 4/1000 = 20/1000 = 1/100 moles. Now on checking for same equivalents in option A: 120 * 0.25/ 1000 = 3/100 moles , hence its not equivalent Option B: 100 * 0.1/1000 = $1 / 100$ moles , it is equivalent Option C : 500 * 0.2/1000 = $1 / 10$ , not equivalent Option D : 10 * 1/1000 = $1 / 100$ moles , equivalent: As asked to opt NOT EQUIVALENT so answer are Option A and C.

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