Question

Sodium bismuthate (NaBiO3) oxidizes MnCl2 to MnO4 and is consequently reduced to Bi(OH)3. How many grams of MnCl2 will be oxidized by 5.0 g of NaBiO3?

Answer 1

The reaction is
14 H+(aq) + 2 Mn2+(aq) + 5 NaBiO3(s) ? 7 H2O(l) + 2 MnO?4(aq) + 5 Bi3+(aq) + 5 Na+(aq)
this tells you that
5 moles NaBiO3 oxidises 2 moles MnCl2

now work out molar masses of each and substitute
5 x molar mass NaBiO3 oxidises 2 molar mass MnCl2
5 x 279.97 g of NaBiO3 oxidises 2 x 125.844 g of MnCl2
1 g of NaBiO3 oxidises (2 x 125.844 / 5 x 279.97) g of MnCl2
1 g of NaBiO3 oxidises 0.18 g of MnCl2
5 g of NaBiO3 oxidises 0.18 * 5 = 0.9 gm of MnCl2