The number of moles of KMnO4 required to oxidise one equivalent of KI in the presence of sulphuric acid is

Question

The number of moles of KMnO4 required to oxidise one equivalent of KI in the presence of sulphuric acid is (A) 5 (B) 2 (C) 1/2 (D) 1/5

Tardigrade Admin · Accepted Answer

overset+7KMnO4+ overset-1KI+H2SO4arrow overset+2MnSO4+ overset0I2+K2SO4+H2O Balance factor for Mn = 5 Balance factor for I=1 Equivalent of KMnO4 = equivalent of KI Number of equivalents = y × Number of moles where, y is the balance factor. =yMn× nMn=yKI× nKI =5× xMn=1×1 nMn=(1/5)

Q. The number of moles of $K M n O_{4}$ required to oxidise one equivalent of $K I$ in the presence of sulphuric acid is

$5$

$2$

$1/2$

$1/5$

Q. The number of moles of KMnO4​ required to oxidise one equivalent of KI in the presence of sulphuric acid is

5

2

1/2

1/5

KMnO4​+7​+KI−1+H2​SO4​→ Mn+2SO4​+I02​+K2​SO4​+H2​O Balance factor for Mn=5 Balance factor for I=1 Equivalent of KMnO4​ = equivalent of KI Number of equivalents = y× Number of moles where, y is the balance factor. =yMn​×nMn​=yKI​×nKI​ =5×xMn​=1×1 nMn​=51​

Q. The number of moles of $K M n O_{4}$ required to oxidise one equivalent of $K I$ in the presence of sulphuric acid is

$5$

$2$

$1/2$

$1/5$