Question

w OIVO SU UGLUUL iz. Demonstrate through mole calculations that iodate was indeed the limiting reactant in the production of triiodide in Part 1 - Goal 3 (titration) of this study. You have three reactants to consider KI, 103, and H. Refer to the titration for Molarities and quantities used for each reactant. Assume that 10 mL of Klau) contains 1 gram of Klin)

#2 please. I have attached info for part 1 goal #3
demonstrate the calculations

Answer 1

Q2) The limiting reactant in a chemical reaction is the substance that is totally consumed when the chemical reaction is completed. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. For a reactant to be consumed completely, it should have number of moles equal to or less than other reactant(s). Therefore, limiting reactant is the reactant with minimum number of moles present.

Now, calculate the number of moles of each reactant and then consider the chemical reaction occuring to determine which reactant is present in least quantity.

Number of moles of Iodate (IO₃^-) present :

In aqueos solution, KIO₃ dissociates as : KIO₃ (aq) $\rightarrow$ K⁺ (aq) + IO₃^- (aq)

Therefore, number of moles of IO₃^- = number of moles of KIO₃ present in solution.

The solution of KIO₃ used is 0.0108M Therefore, 1000ml of solution = 0.0108 moles of KIO₃

10ml of solution = 10 X 0.0108 / 1000 = 0.108 X 10^-3 moles = 0.108mmoles

Therefore, number of moles of IO₃^- = 0.108 mmloes

Number of moles of KI present

10ml of KI solution contains 1gm KI. number of moles = number of grams/ molecular weight

Molecular weight of KI = atomic weight of K + atomic weight of I = 39 + 127 = 166

Therefore, number of moles of KI = 1/166 = 6.024 X 10^-3 moles = 6.024mmoles

Therefore, number of moles of KI = 6.024 mmoles

Now, KI dissociates in aqueous solution as KI (aq.) $\rightarrow$ K⁺ (aq.) + I^- (aq.)   

Therefore, number of moles of I^- present = number of moles of KI = 6.024 moles

Number of moles of H⁺ present

HCl dissociates in aqueous solution as HCl (aq.) $\rightarrow$ H⁺ + Cl^-   

Therefore, Number of moles of H⁺ present = number of moles of HCl present

HCl is 1M. Therefore, 1000ml HCl = 1M HCl

3ml HCl = 3 X 1/1000 = 3X10^-3 moles = 3mmoles

Therefore, number of moles of H⁺ = 3 mmoles

Now, consider the reaction occuring : IO₃^- (aq.) + 8I^- (aq.) + 6H⁺  $\rightarrow$ 3I₃ (aq) + 3H₂O (l)

From this reaction, 1mol IO₃^- (aq) $\equiv$ 8mol I^- (aq.) $\equiv$ 6 mol H⁺ (aq)

$\therefore$ 0.108 moles of IO₃^-$\equiv$ 0.864 moles I^- (aq.) $\equiv$ 0.648 moles H⁺ (aq)

Here, number of moles of I^- present (6.024mmoles) is more than that required to completely convert IO₃^- (0.864 mmoles)

number of moles of H⁺ present (3 mmoles) is more than that required to completely convert IO₃^- (0.648 mmoles)

Therefore, I^-and H⁺ will be present in the reaction mixture after IO₃^- ions have been used up. In other words, amount of the product formed will depend upon IO₃^- because it is the reactant present in minimum quantity and will be completely used up. Therefore IO₃^-​​​​​​​ (iodate) is the limiting reactant.