# Question ac458

Jun 25, 2017

The products are $\text{NaCl} \left(a q\right)$ and ${\text{Fe"_2"S}}_{3} \left(s\right)$

#### Explanation:

We're given a reaction taking place in aqueous solution, and asked to predict the products of the reaction.

For double-replacement reactions like this one, the products can (generally) be found by swapping the cations on each species.

You'll have to keep the oxidation state (charge) the same, so the iron in iron(III) chloride will also have a charge of $3 +$ in the new product.

Swapping the cations on each species, we have

3color(red)("Na")_2"S"(aq) + 2color(blue)("Fe")"Cl"_3(aq) rarr 6color(red)("Na")"Cl" (aq) + overbrace(color(blue)("Fe")_2"S"_3)^"Fe oxidation state +3"(s)#

According to a solubility chart, ${\text{S}}^{2 -}$ is insoluble, and ${\text{Fe}}^{3 +}$ is not an exception, so iron(III) sulfide is the precipitate .

Since not all of the products are present as ions, and a precipitate forms, this reaction does indeed take place. (If the products were all "aqueous", i.e. present as dissolved ions in solution, no reaction would take place, since nothing is really "forming".)