The Iodine "Clock Reaction" Involves the Following Sequence of Reactions $\to$

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) $\to$ 3I₂(aq) + 3H₂O(l)
2) I₂(aq) + 2S₂O₃^2-(aq) $\to$ 2I^-(aq) + S₄O₆^2-(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 (S₂O₃^2-) were needed in reaction 2, in order to react completely with the iodine (I₂) produced in reaction 1?

A) 0) 0020 mol
B) 0) 0030 mol
C) 0) 0040 mol
D) 0) 0050 mol
E) 0) 0060 mol