Lead dioxide oxidises Mn^2+ to permanganate in hot nitric acid: 5PbO2 + 2Mn^2+ + 4H+ -> 2MnO4- + 5Pb^2+ + 2H2O. With Pb = 207 and O = 16, the mass of PbO2 needed to oxidise 0.04 mol of Mn^2+ is

JEE Advanced levelSingle correct optionDifficulty 3 of 3pb218

Lead dioxide oxidises $Mn X^{2 +}$ to permanganate in hot nitric acid: $5 PbO X_{2} + 2 Mn X^{2 +} + 4 H X^{+} 2 MnO X_{4} X^{-} + 5 Pb X^{2 +} + 2 H X_{2} O$ . With Pb = 207 and O = 16, the mass of $PbO X_{2}$ needed to oxidise 0.04 mol of $Mn X^{2 +}$ is

Molar mass of

PbO X_{2}

= 207 + 32 = 239 g/mol. The equation requires 5 mol

PbO X_{2}

per 2 mol

Mn X^{2 +}

, so moles

PbO X_{2}

= 0.04 x 5/2 = 0.10 mol. Mass = 0.10 x 239 = 23.9 g.