Problem 31 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

(a) Precipitation reaction; (b) Precipitation reaction.

Step 1: Write the balanced chemical equation for (a)

To balance the equation, first write the correct formulas with states:\[\text{MnCl}_2(\text{aq}) + \text{Na}_2\text{S}(\text{aq}) \rightarrow \text{MnS}(\text{s}) + 2\text{NaCl}(\text{aq})\]The manganese sulfide (MnS) precipitates out of the solution (solid), so the reaction is completed with one Mn and one S on the product side. Sodium chloride remains aqueous. There are 2 NaCl on the product side to balance Na and Cl atoms.

Step 2: Identify the type of reaction for (a)

This reaction produces a solid precipitate, MnS, from aqueous reactants. Therefore, it is classified as a precipitation reaction.

Step 3: Write the net ionic equation for (a)

First, write the full ionic form of the equation:\[\text{Mn}^{2+}(\text{aq}) + 2\text{Cl}^-(\text{aq}) + 2\text{Na}^+(\text{aq}) + \text{S}^{2-}(\text{aq}) \rightarrow \text{MnS}(\text{s}) + 2\text{Na}^+(\text{aq}) + 2\text{Cl}^-(\text{aq})\]Cancel out the spectator ions, which are \(\text{Na}^+(\text{aq})\) and \(\text{Cl}^-(\text{aq})\), to get the net ionic equation:\[\text{Mn}^{2+}(\text{aq}) + \text{S}^{2-}(\text{aq}) \rightarrow \text{MnS}(\text{s})\]

Step 4: Write the balanced chemical equation for (b)

The balanced equation for part (b) is:\[\text{K}_2\text{CO}_3(\text{aq}) + \text{ZnCl}_2(\text{aq}) \rightarrow \text{ZnCO}_3(\text{s}) + 2\text{KCl}(\text{aq})\]Zinc carbonate forms as a precipitate. Two KCl are needed to balance the equation.

Step 5: Identify the type of reaction for (b)

In this reaction, ZnCO e! precipitates out, thus it is also classified as a precipitation reaction.

Step 6: Write the net ionic equation for (b)

Write the full ionic equation:\[2\text{K}^+(\text{aq}) + \text{CO}_3^{2-}(\text{aq}) + \text{Zn}^{2+}(\text{aq}) + 2\text{Cl}^-(\text{aq}) \rightarrow \text{ZnCO}_3(\text{s}) + 2\text{K}^+(\text{aq}) + 2\text{Cl}^-(\text{aq})\]Remove the spectator ions (\(\text{K}^+\) and \(\text{Cl}^-\)) to get the net ionic equation:\[\text{CO}_3^{2-}(\text{aq}) + \text{Zn}^{2+}(\text{aq}) \rightarrow \text{ZnCO}_3(\text{s})\]

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry, as it ensures that the same number of each type of atom appears on both sides of the equation. This aligns with the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. In simpler terms, whatever you start with must equal what you end with.
When balancing equations, you should follow a methodical approach:

• Begin by writing the unbalanced equation, including the correct chemical formulas for reactants and products.
• Identify the states of each compound, such as solid \( (s) \), liquid \( (\ell) \), gas \( (g) \), or aqueous \( (aq) \).
• Start balancing one element at a time by adjusting coefficients – numbers in front of a compound – not altering subscripts in the formulas.
• Repeat this process until all elements have equal numbers on both sides.

In our example, the equation \( \text{MnCl}_2(\text{aq}) + \text{Na}_2\text{S}(\text{aq}) \rightarrow \text{MnS}(\text{s}) + 2\text{NaCl}(\text{aq}) \) is balanced by having one Mn and S atom on both sides, and two Na and Cl atoms on both sides after balancing.

Net Ionic Equation

Net ionic equations focus on the ions and compounds that are directly involved in the chemical reaction. They help streamline chemical equations by removing "spectator ions," which do not participate in the reaction. To write a net ionic equation, follow these steps:

• Write the full ionic equation by breaking all aqueous compounds into their constituent ions.
• Identify and eliminate spectator ions from both sides of the equation.
• What remains is the net ionic equation, showcasing only the ions involved in forming the solid product.

In our example with \( \text{MnCl}_2(\text{aq}) + \text{Na}_2\text{S}(\text{aq}) \rightarrow \text{MnS}(\text{s}) + 2\text{NaCl}(\text{aq}) \), the ions that form MnS are manganese \( \text{Mn}^{2+}(\text{aq}) \) and sulfide \( \text{S}^{2-}(\text{aq}) \).\ Thus, the net ionic equation is: \( \text{Mn}^{2+}(\text{aq}) + \text{S}^{2-}(\text{aq}) \rightarrow \text{MnS}(\text{s}) \).The spectator ions (Na⁺ and Cl⁻), which remain in solution, are not included in this equation, simplifying our understanding of the underlying chemistry.

Precipitation Reactions

Precipitation reactions are a type of chemical reaction where two aqueous solutions combine to form an insoluble solid, known as a precipitate. They are common in chemistry for identifying and separating different ions. Typically, when two ionic compounds in solution are mixed, they can rearrange to form one or more new compounds.
To determine if a precipitation reaction has occurred:

• Mix two aqueous solutions and observe. Any solid that forms is your precipitate.
• Use solubility rules as a guideline to predict which combinations of ions will result in a precipitate. Some ions generally form soluble compounds, while others typically form solids.
• Write the balanced chemical equation, including states, to clearly show which compound is the precipitate (marked as \( (s) \)).

In our examples, both reactions resulted in precipitates: MnS in the first reaction and ZnCO₃ in the second reaction. The formation of these solids confirms that both reactions are precipitation reactions, and the chemical equations are adjusted accordingly to reflect these solid products.