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:

Step1: Identify the reaction type

This is a double - displacement reaction. The general form of a double - displacement reaction is \(AB + CD
ightarrow AD+CB\), where \(A\) and \(C\) are cations, and \(B\) and \(D\) are anions. For the reaction between \(Fe(C_{2}H_{3}O_{2})_{2}\) (iron(II) acetate) and \(KI\) (potassium iodide), the cations are \(Fe^{2 + }\) and \(K^{+}\), and the anions are \(C_{2}H_{3}O_{2}^{-}\) and \(I^{-}\).

Step2: Predict the products

The possible products of the double - displacement reaction are \(FeI_{2}\) (iron(II) iodide) and \(K C_{2}H_{3}O_{2}\) (potassium acetate).

Step3: Check solubility rules

• Solubility of potassium acetate (\(KC_{2}H_{3}O_{2}\)): Compounds of potassium (\(K^{+}\)) are generally soluble in water. So \(KC_{2}H_{3}O_{2}\) is soluble.
• Solubility of iron(II) iodide (\(FeI_{2}\)): Compounds of iodide (\(I^{-}\)) are generally soluble, except for a few exceptions (like \(AgI\), \(PbI_{2}\), \(Hg_{2}I_{2}\)). Iron(II) iodide (\(FeI_{2}\)) is soluble in water.
• Solubility of reactants: \(Fe(C_{2}H_{3}O_{2})_{2}\) (iron(II) acetate) and \(KI\) (potassium iodide) are both soluble in water (acetates are generally soluble, and potassium salts are generally soluble).

Since both products (\(FeI_{2}\) and \(KC_{2}H_{3}O_{2}\)) are soluble in water, no precipitate will form.

Answer:

D. none