Sodium iodate reacts with sodium sulfite according to the equation :
NaIO3 + 3Na2SO3 → 3Na2SO4 + NaI
a). In this reaction, which substance is the oxidising agent? Why?
b). How many grams of NaIO3 are needed to react with 5.00g of Na2SO3?
a. NaIO3 is the Oxidizing agent.
The reason is that the Oxidation number of Iodine changes from +5 in NaIO3 to -1 in NaI. This shows that the compound itself undergoes reduction (as an oxidizing agent)
b. Molar mass of NaIO3= 198g/mol
Molar mass of Na2SO3= 86g/mol
From the equation
3(86)g of Na2SO3 reacts with 198g of NaIO3
5.00g of Na2SO3 will react with
198 x 5.00/258 = 3.84g of NaIO3