Question

I think my main question/concern really relates to the formation of a product that is listed as reacting with water after the combination of two aqueous solutions.

For example, both sodium iodide (NaI) and iron(III) chloride (FeCl₃) form aqueous solutions. When combined, the expected products from double replacement would be sodium chloride (NaCl) and iron(III) iodide (FeI₃). Equation something like this (Where the ? indicates the root of my question):

3NaI(aq)+FeCl₃(aq)⟶3NaCl(aq)+FeI₃ (?)

Now, FeI₃ is listed as reacting with water for its solubility. Does this mean that FeI₃ is, in fact, not produced, but rather begins decomposing into iron(II) iodide (FeI₂) and elemental iodine (I₂)? Or do we get some form of iron compound with an iodine oxyanion like Fe(IO₃)₃ and hydrogen gas (H₂)?

And in either case, would this reaction be said to take place or not? I'm not really sure what my end goal is here, I just don't understand how an ionic compound that might be formed from aqueous solutions will behave in reacting with water considering its surrounded by water from the two aqueous solutions.

Thanks in advance for the help!

Answer 1

In a double replacement reaction between sodium iodide and iron(III) chloride, sodium chloride and iron(III) iodide are the expected products. However, iron(III) iodide is not stable in water and decomposes into iron(II) iodide and elemental iodine. This reaction takes place because iron(II) iodide is insoluble in water.

Step-by-step explanation:

When sodium iodide (NaI) and iron(III) chloride (FeCl₃) react in a double replacement reaction, the expected products are sodium chloride (NaCl) and iron(III) iodide (FeI₃). However, FeI₃ is not stable in water and undergoes decomposition. It forms iron(II) iodide (FeI₂) and elemental iodine (I₂). So, the reaction can be represented as:

3NaI(aq) + FeCl₃(aq) → 3NaCl(aq) + FeI₂(s) + I₂(g)

This reaction takes place because FeI₂ is insoluble in water and precipitates out of the solution.