The iodine “clock reaction” involves the following sequence of reactions occurri

Question

The iodine “clock reaction” involves the following sequence of reactions occurring in a reaction mixture in a single beaker.
1. IO3-(aq) + 5I-(aq) + 6H+(aq) ? 3I2(aq) + 3H2O(l)
2. I2(aq) + 2S2O32-(aq) ? 2I-(aq) + S4O62-(aq)
The molecular iodine (I2) formed in reaction 1 is immediately used up in reaction 2, so that no iodine accumulates. In one experiment, a student made up a reaction mixture which initially contained 0.0020 mol of iodate ions (IO3-). If the iodate ions reacted completely, how many moles of thiosulfate ions (S2O32-) were needed in reaction 2, in order to react completely with the iodine (I2) produced in reaction 1?

Explanation / Answer

IO3-(aq) + 5I-(aq) + 6H+(aq) = 3I2(aq) + 3H2O(l)

From the stoichiometry of the above reaction

1 mol IO3- produces = 3 mol I2

0.0020 mol IO3- produces = 3*0.0020 = 0.0060 mol I2

I2(aq) + 2S2O32-(aq) ? 2I-(aq) + S4O62-(aq)

From the stoichiometry of the reaction 2

1 mol I2 reacts with = 2 mol S2O3 2-

0.0060 mol I2 reacts with = 2*0.0060

= 0.0120 mol S2O3 2-

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