In dilute nitric acid, Mn^2+ is oxidised by sodium bismuthate to MnO4-, the bismuth being reduced from Bi(V) to Bi^3+. What mass of NaBiO3 (molar mass 280 g mol^-1) is required to oxidise 0.0120 mol of Mn^2+ completely?

JEE Advanced levelSingle correct optionDifficulty 3 of 3pn013

In dilute nitric acid, $Mn X^{2 +}$ is oxidised by sodium bismuthate to $MnO X_{4} X^{-}$ , the bismuth being reduced from $Bi (V)$ to $Bi X^{3 +}$ . What mass of $NaBiO X_{3}$ (molar mass $280 g mo l^{- 1}$ ) is required to oxidise $0.0120$ mol of $Mn X^{2 +}$ completely?

Mn goes from +2 to +7 (5 e

^{-}

) and Bi from +5 to +3 (2 e

^{-}

), so

2 Mn X^{2 +}

react with

5 NaBiO X_{3}

, i.e. a mole ratio of 2.5 per Mn. Moles of

NaBiO X_{3}

0.0120 \times 2.5 = 0.0300

mol. Mass =

0.0300 \times 280 = 8.40 g