Question

write the balanced net ionic equation for the reaction when magnesium iodide and sodium carbonate are mixed in aqueous solution. if no reaction occurs, simply write only nr. be sure to include the proper phases for all species within the reaction.

Explanation:

Step1: Write the formulas of reactants

Magnesium iodide: \(MgI_2\), sodium carbonate: \(Na_2CO_3\). Both are in aqueous solution, so \(MgI_2(aq)\) and \(Na_2CO_3(aq)\)

Step2: Predict the products

Using double - displacement reaction rules (\(AB + CD
ightarrow AD+CB\)), the products are \(MgCO_3\) and \(NaI\). \(MgCO_3\) is insoluble (from solubility rules: carbonates of group 2 metals are insoluble), so \(MgCO_3(s)\), and \(NaI(aq)\)

Step3: Write the molecular equation

\(MgI_2(aq)+Na_2CO_3(aq)
ightarrow MgCO_3(s) + 2NaI(aq)\)

Step4: Write the total ionic equation

\(Mg^{2 +}(aq)+2I^{-}(aq)+2Na^{+}(aq)+CO_{3}^{2 -}(aq)
ightarrow MgCO_3(s)+2Na^{+}(aq)+2I^{-}(aq)\)

Step5: Write the net ionic equation

Cancel out the spectator ions (\(Na^{+}\) and \(I^{-}\))
\(Mg^{2 +}(aq)+CO_{3}^{2 -}(aq)
ightarrow MgCO_3(s)\)

Answer:

\(Mg^{2 +}(aq)+CO_{3}^{2 -}(aq)
ightarrow MgCO_3(s)\)