Question

select the correct answer
which compound will precipitate when solutions of iron(ii) acetate, fe(c₂h₃o₂)₂, and potassium iodide (ki) are mixed?
a. fe(c₂h₃o₂)₂
b. fei₂
c. kch₃co₂
d. none

Explanation:

To solve this, we analyze each option using solubility rules:

Step 1: Analyze Option A (\( \text{Fe}(\text{C}_2\text{H}_3\text{O}_2)_2 \))

Iron(II) acetate is one of the reactants. No precipitation occurs here (it’s the starting compound).

Step 2: Analyze Option B (\( \text{FeI}_2 \))

From solubility rule 3: Chlorides, bromides, iodides (\( \text{I}^- \)) are soluble, except for \( \text{Ag}^+ \), \( \text{Hg}_2^{2+} \), \( \text{Pb}^{2+} \). Iron(II) (\( \text{Fe}^{2+} \)) is not in the “insoluble” exception list for iodides. Thus, \( \text{FeI}_2 \) is soluble.

Step 3: Analyze Option C (\( \text{KCH}_3\text{CO}_2 \))

From solubility rule 2: Acetates (\( \text{C}_2\text{H}_3\text{O}_2^- \)) are soluble. Potassium (\( \text{K}^+ \)) is a Group 1 metal (rule 1: Group 1 compounds are soluble). Thus, potassium acetate is soluble.

Step 4: Analyze Option D (None)

Since \( \text{FeI}_2 \), \( \text{KCH}_3\text{CO}_2 \), and \( \text{Fe}(\text{C}_2\text{H}_3\text{O}_2)_2 \) are all soluble, no precipitate forms.

Answer:

D. none