Question

Potassium Iodate solution was prepared by dissolving 10.22 of KIO3 in a 500 mL volumetric flask. Then 50 mL of the solution was pipetted into a flask and treated with KI (1g) and acid (10 mL of 0.5 H2SO4) How many moles of I3 are created by the reaction. The answer is 2.26 just please show the work.

Answer 1

Moles of KIO₃ taken = Mass/MW = 10.22/241 = 0.0424

Since only 50 mL of the solution is taken, so moles of KIO₃ taken for reaction = 0.0424/10 = 0.00424 moles

Moles of KI taken = 1/166 = 0.00602

The reaction taking place is:

IO₃^- + 8I^- + 6H⁺ ---> 3I₃^- + 3H₂O

According to the reaction stoichiometry, KI is the limiting reagent.

So,

Moles of I₃ produced = (3/8)*Moles of I^- present = (3/8)*0.00602 = 0.0022575 moles = 2.26 millimoles

The answer provided by you ( 2.26 moles) is a very big number and is not possible at all. Please verify.