In the bismuthate test, 0.020 mol of Mn^2+ is completely oxidised to MnO4- in nitric acid by solid NaBiO3 (Bi goes from +5 to +3). The minimum mass of NaBiO3 required is (Na 23, Bi 209, O 16)

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In the bismuthate test, 0.020 mol of $Mn X^{2 +}$ is completely oxidised to $MnO X_{4} X^{-}$ in nitric acid by solid $NaBiO X_{3}$ ( $Bi$ goes from $+ 5$ to $+ 3$ ). The minimum mass of $NaBiO X_{3}$ required is (Na 23, Bi 209, O 16)

Mn X^{2 +}

loses 5 electrons (

+ 2

+ 7

) and each

NaBiO X_{3}

gains 2, so the ratio is

2 Mn X^{2 +}

5 NaBiO X_{3}

. Moles of

NaBiO X_{3}

= 0.020 x 5/2 = 0.050 mol, and the molar mass is 23+209+48 = 280 g/mol. Mass = 0.050 x 280 = 14.0 g.