Problem 61 - Chemistry and Chemical Reactivity

Question

Classify each of the following reactions as a precipitation, acid-base, or gas-forming reaction. Show states for the products \((s, \ell, g, a q),\) and then balance the completed equation. Write the net ionic equation. (a) \(\mathrm{MnCl}_{2}(\mathrm{aq})+\mathrm{Na}_{2} \mathrm{S}(\mathrm{aq}) \rightarrow \mathrm{MnS}+\mathrm{NaCl}\) (b) \(\mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{aq})+\mathrm{ZnCl}_{2}(\mathrm{aq}) \rightarrow \mathrm{ZnCO}_{3}+\mathrm{KCl}\)

Step-by-Step Solution

Answer

Both reactions (a) and (b) are precipitation reactions with their respective net ionic equations: (a) \(\mathrm{Mn}^{2+}(\mathrm{aq}) + \mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{MnS}(s)\). (b) \(\mathrm{Zn}^{2+}(\mathrm{aq}) + \mathrm{CO}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{ZnCO}_{3}(s)\).\n

Step 1: Identify reaction type for (a)

Observe that MnS is an insoluble salt that precipitates out of the solution. Hence, the reaction \(\mathrm{MnCl}_{2}(\mathrm{aq})+\mathrm{Na}_{2} \mathrm{S}(\mathrm{aq}) \rightarrow \mathrm{MnS}+\mathrm{NaCl}\) is a precipitation reaction.

Step 2: Write full equation with states for (a)

The reaction products need states indicated: \[\mathrm{MnCl}_{2}(\mathrm{aq}) + \mathrm{Na}_{2} \mathrm{S}(\mathrm{aq}) \rightarrow \mathrm{MnS}(s) + 2\mathrm{NaCl}(\mathrm{aq})\] The balancing has been done by adjusting coefficients to match the number of atoms for each element on both sides.

Step 3: Write net ionic equation for (a)

Remove the spectator ions \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) that remain in the solution: \[\mathrm{Mn}^{2+}(\mathrm{aq}) + \mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{MnS}(s)\]

Step 4: Identify reaction type for (b)

ZnCO\(_3\) is insoluble and precipitates out of the solution, indicating that \(\mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{aq})+\mathrm{ZnCl}_{2}(\mathrm{aq}) \rightarrow \mathrm{ZnCO}_{3}+\mathrm{KCl}\) is a precipitation reaction.

Step 5: Write full equation with states for (b)

The fully identified reaction: \[\mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{aq}) + \mathrm{ZnCl}_{2}(\mathrm{aq}) \rightarrow \mathrm{ZnCO}_{3}(s) + 2\mathrm{KCl}(\mathrm{aq})\]\Balance the equation by ensuring the same number of atoms on both sides.

Step 6: Write net ionic equation for (b)

The spectator ions \(\mathrm{K}^{+}\) and \(\mathrm{Cl}^{-}\) are removed, yielding: \[\mathrm{Zn}^{2+}(\mathrm{aq}) + \mathrm{CO}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{ZnCO}_{3}(s)\]

Key Concepts

Precipitation Reaction

In chemistry, a precipitation reaction occurs when two soluble reactants in aqueous solutions form an insoluble solid product, known as a precipitate. This reaction is a type of double displacement reaction where the cations and anions of the reactants exchange places to form a new compound. The formation of a solid from a solution is a critical indication that a precipitation reaction has occurred.
For example, consider the reaction between manganese(II) chloride (\( \mathrm{MnCl}_{2}(aq) \)) and sodium sulfide (\( \mathrm{Na}_{2} \mathrm{S}(aq) \)). When these two solutions are mixed, the product is manganese(II) sulfide (\( \mathrm{MnS}(s) \)), which is insoluble in water and forms a solid, or precipitate. The chemical equation for this reaction can be written as:

• \( \mathrm{MnCl}_{2}(aq) + \mathrm{Na}_{2} \mathrm{S}(aq) \rightarrow \mathrm{MnS}(s) + 2\mathrm{NaCl}(aq) \)

A solid precipitate, such as \( \mathrm{MnS} \), indicates the occurrence of a precipitation reaction.

Net Ionic Equations

Net ionic equations focus on the ions involved in the actual chemical change, omitting the spectator ions that do not participate directly in the reaction. This approach highlights the essence of the chemical process by stripping away the ions that merely exist in the solution but do not affect the overall reaction.
For example, in the precipitation of manganese(II) sulfide from the solutions of manganese chloride and sodium sulfide, the sodium \( \mathrm{Na}^{+} \) and chloride ions \( \mathrm{Cl}^{-} \) remain in the aqueous phase and do not contribute to the formation of the precipitate. As a result, the net ionic equation solely involves the combination of manganese(II) ions and sulfide ions to form manganese sulfide:

• \( \mathrm{Mn}^{2+}(aq) + \mathrm{S}^{2-}(aq) \rightarrow \mathrm{MnS}(s) \)

Net ionic equations are important because they simplify reactions, showing only the substances that undergo change and emphasizing the main reaction occurring.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures the conservation of mass and conforming to the law that matter is neither created nor destroyed. Each side of the chemical equation must represent the same quantity of each type of atom.
In a chemical equation, we balance the equation by adjusting the coefficients before the reactants and products. For instance, in the reaction:

• \( \mathrm{K}_{2} \mathrm{CO}_{3}(aq) + \mathrm{ZnCl}_{2}(aq) \rightarrow \mathrm{ZnCO}_{3}(s) + 2\mathrm{KCl}(aq) \)

The coefficients ensure the number of potassium \( \mathrm{K} \) atoms, carbonate \( \mathrm{CO}_{3} \), and chloride \( \mathrm{Cl} \) ions are equal on both the reactant and product sides of the equation. In essence, balancing confirms the chemical equation reflects the reality of the matter involved in the chemical transformation.