The number of moles of H-2-O-2- needed to reduce 1 mole of KMnO-4- in acidic medium are-2 moles5 moles3 moles2-5 moles

Question

Toppr Admin · Accepted Answer

The overall reaction is- 6H-2MnO-x2212-4-5H2O2-x2192-5O2-2Mn2-8H2O-Therefore- the valency factor of H2O2 is 2- Since- the valency factor of KMnO4 is 5 in acidic medium- the number of moles required to reduce is 52 moles-

The number of moles of $H_{2} O_{2}$ needed to reduce $1$ mole of $K M n O_{4}$ in acidic medium are:
$2$ moles
$5$ moles
$3$ moles
$2.5$ moles

The overall reaction is, $6 H^{+} + 2 M n O_{4}^{-} + 5 H_{2} O_{2} \to 5 O_{2} + 2 M n^{2 +} + 8 H_{2} O$ .

Therefore, the valency factor of $H_{2} O_{2}$ is $2$ . Since, the valency factor of $K M n O_{4}$ is $5$ in acidic medium, the number of moles required to reduce is $\frac{5}{2}$ moles.

The number of moles of H2O2 needed to reduce 1 mole of KMnO4 in acidic medium are:2 moles5 moles3 moles2.5 moles

The overall reaction is, 6H++2MnO−4+5H2O2→5O2+2Mn2++8H2O.Therefore, the valency factor of H2O2 is 2. Since, the valency factor of KMnO4 is 5 in acidic medium, the number of moles required to reduce is 52 moles.

The number of moles of $H_{2} O_{2}$ needed to reduce $1$ mole of $K M n O_{4}$ in acidic medium are:
$2$ moles
$5$ moles
$3$ moles
$2.5$ moles

The overall reaction is, $6 H^{+} + 2 M n O_{4}^{-} + 5 H_{2} O_{2} \to 5 O_{2} + 2 M n^{2 +} + 8 H_{2} O$ .

Therefore, the valency factor of $H_{2} O_{2}$ is $2$ . Since, the valency factor of $K M n O_{4}$ is $5$ in acidic medium, the number of moles required to reduce is $\frac{5}{2}$ moles.