KMnO4 reacts with oxalic acid according to the equation2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O,20ml of 0.1 M KMnO4 will react with

Equation of KMnO4=20×0.1×5=10Equation of oxalic acid =25×0.2×2=10

KMnO4 reacts with oxalic acid according to the equation2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O,20ml of 0.1 M KMnO4 will react with

Equation of KMnO4=20×0.1×5=10Equation of oxalic acid =25×0.2×2=10

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

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120 mL of 0.25 $M H_{2} C_{2} O_{4}$

150 mL of 0.10 $M H_{2} C_{2} O_{4}$

25 mL of 0.20 $M H_{2} C_{2} O_{4}$

50 mL of 0.20 $M H_{2} C_{2} O_{4}$

answer is C.

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Equation of $K M n O_{4} = 20 \times 0.1 \times 5 = 10$

Equation of oxalic acid $= 25 \times 0.2 \times 2 = 10$

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KMnO4 reacts with oxalic acid according to the equation2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O,20ml of 0.1 M KMnO4 will react with

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