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Stoichiometry

KMnO 4 reacts...

Question

$K M n O_{4}$ reacts with oxalic acid according to the equation,
$2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$
Here, 20 mL of 0.1 M $K M n O_{4}$ is equivalent to:

20 mL of 0.5 M

H_{2} C_{2} O_{4}

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50 mL of 0.1 M

H_{2} C_{2} O_{4}

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50 ml of 0.01 M

H_{2} C_{2} O_{4}

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20 ml of 0.1 M

H_{2} C_{2} O_{4}

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Solution

The correct option is B 50 mL of 0.1 M $H_{2} C_{2} O_{4}$
50 mL of 0.1 M $H_{2} C_{2} O_{4}$
Equivalents of $M n O^{4 -}$ = Equivalents of $C_{2} O_{4}^{2 -}$
$M \times V \times n_{f} = M \times V \times n_{f}$
$(0.1) \times (20) \times 5 = (50) \times (0.1) \times 2$
$10 = 10$

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