Question

Demonstrate through mole calculations that iodate was indeed the limiting reactant in the production of triiodide in Part I – Goal 3 (titration) of this study. You have three reactants to consider KI, IO3 - , and H+ . Refer to the titration for Molarities and quantities used for each reactant. Assume that 10 mL of KI(aq) contains 1 gram of KI(s).

Answer 1

IO₃^- (aq) + 8I^-(aq) + 6H⁺ (aq) --------> 3I₃^-(aq) + 3 H₂0(l)
to prove that iodate is indeed the limiting reactant, we need to calculate individual
number of moles of each reactant. But looking at the reaction stoichiometry itself
we can say that IO₃^- is limiting reactant because 1 mol of IO₃^- reacts to give away
3 mol of of I₃^- , whereas the 8 KI ions produce 3 mol of triiodide and 6 mol of H⁺
produce 3 mol of triiodide. A limiting reactant has least number of moles when compared
to other reactant and it will run out first.