A student mixed the following reactants, using the general procedure for the exp

Question

A student mixed the following reactants, using the general procedure for the experiment: 10.0 mL, 0.10 M KI, 10.0 mL 0.0010 M Na_S_2 O_3, 20.0 ml, 0.040 M KBrO_3, and 10.0 ml, 0.10 M HCl. It took about 75 seconds for the mixture to turn blue. By using the dilution equation calculate the concentrations of each reactant in the mixture: [I^-] = [BrO_3^-] = [H^+] = Calculate the moles of S_20_3^2- consumed by reaction with I_2 within the 75 s it took for the mixture to turn blue: Calculate the rate of the reaction delta [BrO_3^-]/delta t, by using the stoichiometry of the reaction and the total volume: Rate = The student then prepared Reaction Mixture 1, as described in the Part 1 of the Procedure, and it required about 155 seconds to change color.

Explanation / Answer

Total volume of the reactant mixture = 50 ml = 0.05 litres

moles = molar concentration*volume of solution in litres

KI(aq) ------> K+(aq) + I-(aq)

thus, [I-] = [K+] = [KI] = 0.1*0.01/0.05 = 0.02 M

KBrO3(aq) ------> K+(aq) + BrO3-(aq)

thus, [BrO3-] = [K+] = [KBrO3] = 0.02*0.04/0.05 = 0.016 M

HCl(aq) ------> H+(aq) + Cl-(aq)

[H+] = [Cl-] = [HCl] = 0.1*0.01/0.05 = 0.02 M

overall reaction :-

6KI + 6HCl + KBrO3 ------> 3I2 + KBr + 6KCl + 3H2O

moles of I2 produced = (1/2)*moles of KI reacting = 0.5*0.02*0.05 = 0.0005

I2 + 2Na2S2O3 ----> Na2S4O6 + 2NaI

Thus, moles of Na2S2O3 consumed = 2*moles of I2 formed = 0.001

But Na2S2O3 is only 0.00002 moles

Thus, S2O32- consumed = 0.00002 moles

Now, KBrO3 consumed = (1/6)*moles of KI reacting = 0.001

Initial rate of reaction:- = change in concentration of KBrO3/time interval = (0.016 - 0.001)/75 = 0.0002 M/s

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