Acidified KMnO4 oxidises Fe^2+ to Fe^3+ and is itself reduced to Mn^2+. Calculate the mass of KMnO4 required to oxidise 11.2 g of Fe^2+ ions completely. Atomic masses: Fe 56, K 39, Mn 55, O 16. Give the answer in grams to two decimal places.

JEE Main levelNumerical answerDifficulty 2 of 3rx087

Acidified $KMnO X_{4}$ oxidises $Fe X^{2 +}$ to $Fe X^{3 +}$ and is itself reduced to $Mn X^{2 +}$ . Calculate the mass of $KMnO X_{4}$ required to oxidise 11.2 g of $Fe X^{2 +}$ ions completely. Atomic masses: Fe 56, K 39, Mn 55, O 16. Give the answer in grams to two decimal places.

Numerical answer type, answer in g, tolerance 0.02.

The ion-electron method gives

MnO X_{4} X^{-} + 5 Fe X^{2 +} + 8 H X^{+} Mn X^{2 +} + 5 Fe X^{3 +} + 4 H X_{2} O

. Moles of

Fe X^{2 +}

= 11.2/56 = 0.20 mol, so

MnO X_{4} X^{-}

needed = 0.20/5 = 0.04 mol. Molar mass of

KMnO X_{4}

= 39 + 55 + 64 = 158 g/mol, so mass = 0.04 x 158 = 6.32 g.