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When we have two test tubes and we add:

  1. In the first test tube: hexane, aqueous solution of $\ce{NaCl}$ and aqueous solution of $\ce{I2}$

  2. In the second test tube: hexane, aqueous solution of $\ce{NaI}$ and aqueous solution of $\ce{Cl2}$

We are asked to write the reactions that take place in these experiments. Do we write only this:

$$\ce{NaCl + I2 → no reaction}$$

$$\ce{2NaI + Cl2 → 2NaCl + I2 (displacement)}$$

Or should hexane appear in the reactions too?

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    $\begingroup$ I'd write the reactions the same way you did. What puzzles me is why they bothered to mention hexane at all. $\endgroup$ Commented Mar 9, 2017 at 18:02

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Hexane is the solvent for the two halogens. It does not react with either so leave it out of your answer.

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