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? Multiple Choice 0.0020 mol 0.012 mol 0.0040 mol

Answer 1

C: 0.012 mol.

Explanation

Start with 0.0020 moles of iodate ions ${\text{IO}_{3}}^{-}$.

How many moles of iodine $\text{I}_2$ will be produced?

${\text{IO}_{3}}^{-}$ converts to $\text{I}_2$ in the first reaction. The coefficient in front of $\text{I}_2$ is three times the coefficient in front of ${\text{IO}_{3}}^{-}$. In other words, each mole of ${\text{IO}_{3}}^{-}$ will produce three moles of $\text{I}_2$. 0.0020 moles of ${\text{IO}_{3}}^{-}$ will convert to 0.0060 moles of $\text{I}_2$.

How many moles of thiosulfate ions ${\text{S}_2\text{O}_3}^{2-}$ are required?

$\text{I}_2$ reacts with ${\text{S}_2\text{O}_3}^{2-}$ in the second reaction. The coefficient in front of $\text{I}_2$ is twice the coefficient in front of ${\text{S}_2\text{O}_3}^{2-}$. How many moles of ${\text{S}_2\text{O}_3}^{2-}$ does each mole of $\text{I}_2$ consume? Two. 0.0060 moles of $\text{I}_2$ will be produced. As a result, $2 \times 0.0060 = 0.0120$ moles of ${\text{S}_2\text{O}_3}^{2-}$ will be needed.