Question

2. 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, iodate, and H+. Assume that 1/4 teaspoon of KI is approximately 1 g. Im not sure how to find the products to do the work with this equation. Also, how do I know how the coefficients of the reactants. A molarity of 0.0108 was used for IO3. I know how to do limiting reactant problems, however I am lost on this problem.

Answer 1

Solution.

The amount of iodate-ions is:

$n=cV = 0.0108\times 0.050 = 5.4\times 10^{-4} \ moles;$

The mass of KI is 4 g as 1 teaspoon is 4 g. The molar mass of KI is 166 g/mol.

The amount of KI is

$n=\frac{m}{M} = \frac{4}{166}=0.024 \ moles;$

According to the stoichiometry of the reaction

IO₃^- (aq) + 8I^-(aq) + 6 H⁺(aq) → 3I₃^-(aq) + 3H₂O(l)

1 mole of IO₃^- (aq) requires 8 moles of I^-(aq) ;

0.00054  moles of IO₃^- (aq) requires x moles of I^-(aq) ;

$x = 8\times 0.00054 = 4.32\times 10^{-3} \ moles;$

It means that 0.0043 moles of KI would be enough. But we have 0.024 moles, so this reactant will remain after the reaction end. Therefore, the iodate-ion is a limiting reactant.