KMnO4 reacts with oxalic acid according to the equation: 2MnO-4+5C2O2-4+16H+→2Mn2++10CO2+8H2O Here 20 ml of 0.1 M KMnO4 is equivalent to

Question

KMnO4 reacts with oxalic acid according to the equation: 2MnO-4+5C2O2-4+16H+→2Mn2++10CO2+8H2O Here 20 ml of 0.1 M KMnO4 is equivalent to (A) 20 ml of 0.5 M C2H2O4 (B) 50 ml of 0.1 MC2H2O4 (C) 20 ml of 0.5 M C2H2O4 (D) 20 ml of 0.1 MC2H2O4

Tardigrade Admin · Accepted Answer

Correct answer is (b) 50 ml of 0.1 MC2H2O4

Q. $K M n O_{4}$ 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 $20 m l$ of $0.1 M K M n O_{4}$ is equivalent to

$20 m l$ of $0.5 M C_{2} H_{2} O_{4}$

$50 m l$ of $0.1 M C_{2} H_{2} O_{4}$

$20 m l$ of $0.5 M C_{2} H_{2} O_{4}$

$20 m l$ of $0.1 M C_{2} H_{2} O_{4}$

Correct answer is (b) $50 m l$ of $0.1 M C_{2} H_{2} O_{4}$

Q. KMnO4​ reacts with oxalic acid according to the equation: 2MnO4−​+5C2​O42−​+16H+→2Mn2++10CO2​+8H2​O Here 20ml of 0.1MKMnO4​ is equivalent to

20ml of 0.5MC2​H2​O4​

50mlof 0.1MC2​H2​O4​

20mlof 0.5MC2​H2​O4​

20mlof 0.1MC2​H2​O4​

Correct answer is (b) 50mlof 0.1MC2​H2​O4​

Q. $K M n O_{4}$ 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 $20 m l$ of $0.1 M K M n O_{4}$ is equivalent to

$20 m l$ of $0.5 M C_{2} H_{2} O_{4}$

$50 m l$ of $0.1 M C_{2} H_{2} O_{4}$

$20 m l$ of $0.5 M C_{2} H_{2} O_{4}$

$20 m l$ of $0.1 M C_{2} H_{2} O_{4}$

Correct answer is (b) $50 m l$ of $0.1 M C_{2} H_{2} O_{4}$