Chemistry Forum

Triodide anions are generated in solution by the following reaction in acidic solution:
IO₃^- + 8I^- + 6H⁺ = 3I₃^- + 3H₂O balanced

I am trying to solve the below problem.
A sample of 0.6013 g of potassium iodate was dissolved in water. Hydrochloric acid and solid potassium iodide were then added in excess. What is the minimum mass of of solid KI and the minimum volume of 3.00 M HCl required to convert all the ions IO₃– to I₃– ions?

Could you help me to figure out how to write a balanced equation and how to approach this problem.

Thank you,

Thomas wrote:Triodide anions are generated in solution by the following reaction in acidic solution:
IO₃^- + 8I^- + 6H⁺ = 3I₃^- + 3H₂O balanced

I am trying to solve the below problem.
A sample of 0.6013 g of potassium iodate was dissolved in water. Hydrochloric acid and solid potassium iodide were then added in excess. What is the minimum mass of of solid KI and the minimum volume of 3.00 M HCl required to convert all the ions IO₃– to I₃– ions?

Could you help me to figure out how to write a balanced equation and how to approach this problem.

Thank you,

Since you have already balanced net ionic equation, you need just to substitute reagent ions to the relevant compounds and you will get balanced molecular equation:
KIO₃ + 8KI + 6HCl = 3KI₃ + 3H₂O + 6KCl
Uou know that KI and HCl are in excess, thus KIO₃ is the limiting regeant and number of moles of other compounds should be calculated from the number of moles of KIO₃.
Does that clarifies this problem?