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Chemistry

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baherus [9]3 years ago

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Alexxx [7]3 years ago

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The iodine "clock reaction" involves the following sequence of reactions occurring in a reaction mixture in a single beaker. 1.

Mars2501 [29]

C: 0.012 mol.

<h3>Explanation</h3>

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.