Problem 53 - Chemistry and Chemical Reactivity

Question

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

Step 1: Writing the Unbalanced Equation for (a)

The initial unbalanced chemical equation is:\[ \mathrm{MnCl}_{2}(\mathrm{aq}) + \mathrm{Na}_{2} \mathrm{S}(\mathrm{aq}) \rightarrow \mathrm{MnS} + \mathrm{NaCl} \]

Step 2: Determine Physical States for (a)

\(\mathrm{MnS}\) is a solid (precipitate), while \(\mathrm{NaCl}\) remains in solution as aqueous:\[ \mathrm{MnS}(\mathrm{s}) + \mathrm{NaCl}(\mathrm{aq}) \]

Step 3: Balance the Equation for (a)

The balanced equation is:\[ \mathrm{MnCl}_{2}(\mathrm{aq}) + \mathrm{Na}_{2} \mathrm{S}(\mathrm{aq}) \rightarrow \mathrm{MnS}(\mathrm{s}) + 2\mathrm{NaCl}(\mathrm{aq}) \]

Step 4: Write Net Ionic Equation for (a)

Start by writing the complete ionic equation:\[ \mathrm{Mn}^{2+}(\mathrm{aq}) + 2\mathrm{Cl}^{-}(\mathrm{aq}) + 2\mathrm{Na}^{+}(\mathrm{aq}) + \mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{MnS}(\mathrm{s}) + 2\mathrm{Na}^{+}(\mathrm{aq}) + 2\mathrm{Cl}^{-}(\mathrm{aq}) \]Cancel spectator ions (\(\mathrm{Na}^{+}(\mathrm{aq})\) and \(\mathrm{Cl}^{-}(\mathrm{aq})\)):\[ \mathrm{Mn}^{2+}(\mathrm{aq}) + \mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{MnS}(\mathrm{s}) \]

Step 5: Classify Reaction (a)

This reaction is a precipitation reaction because it forms a solid precipitate, \(\mathrm{MnS}(\mathrm{s})\).

Step 6: Writing the Unbalanced Equation for (b)

The initial unbalanced chemical equation is:\[ \mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{aq}) + \mathrm{ZnCl}_{2}(\mathrm{aq}) \rightarrow \mathrm{ZnCO}_{3} + \mathrm{KCl} \]

Step 7: Determine Physical States for (b)

\(\mathrm{ZnCO}_{3}\) is a solid (precipitate), while \(\mathrm{KCl}\) remains in solution as aqueous:\[ \mathrm{ZnCO}_{3}(\mathrm{s}) + \mathrm{KCl}(\mathrm{aq}) \]

Step 8: Balance the Equation for (b)

The balanced equation is:\[ \mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{aq}) + \mathrm{ZnCl}_{2}(\mathrm{aq}) \rightarrow \mathrm{ZnCO}_{3}(\mathrm{s}) + 2\mathrm{KCl}(\mathrm{aq}) \]

Step 9: Write Net Ionic Equation for (b)

Start by writing the complete ionic equation:\[ 2\mathrm{K}^{+}(\mathrm{aq}) + \mathrm{CO}_{3}^{2-}(\mathrm{aq}) + \mathrm{Zn}^{2+}(\mathrm{aq}) + 2\mathrm{Cl}^{-}(\mathrm{aq}) \rightarrow \mathrm{ZnCO}_{3}(\mathrm{s}) + 2\mathrm{K}^{+}(\mathrm{aq}) + 2\mathrm{Cl}^{-}(\mathrm{aq}) \]Cancel spectator ions (\(\mathrm{K}^{+}(\mathrm{aq})\) and \(\mathrm{Cl}^{-}(\mathrm{aq})\)):\[ \mathrm{Zn}^{2+}(\mathrm{aq}) + \mathrm{CO}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{ZnCO}_{3}(\mathrm{s}) \]

Step 10: Classify Reaction (b)

This reaction is a precipitation reaction because it forms a solid precipitate, \(\mathrm{ZnCO}_{3}(\mathrm{s})\).

Key Concepts

Precipitation Reaction

A precipitation reaction occurs when two aqueous solutions combine and form an insoluble solid, known as a precipitate. This type of reaction is typically seen in double displacement reactions where two ions bond, forming the solid precipitate. For instance, in the reaction between manganese chloride (MnCl₂) and sodium sulfide (Na₂S), a solid precipitate, manganese sulfide (MnS), forms and settles out of the solution.
To recognize a precipitation reaction, look for evidence of a solid forming in the mixed solutions. The resulting precipitate is often visible as a cloudiness or solid "flakes" in the liquid.

• Visual cue: formation of a solid precipitate
• Reactants: typically aqueous ionic compounds
• Resulting solid: often written with (s) to indicate its state

This process is useful in various applications, including the removal of contaminants in water treatment or in qualitative chemical analysis to identify the presence of certain ions.

Balancing Equations

Balancing chemical equations is crucial to satisfying the law of conservation of mass, which states that matter cannot be created or destroyed in a closed system. This means the number of atoms for each element must be equal on both sides of the chemical equation.
To balance the equation, follow these general steps:

• Identify the reactants and products, write their chemical formulas.
• Count the number of atoms of each element on both sides of the equation.
• Use coefficients to balance the number of atoms for each element.
• Recheck to ensure all atoms balance and adjust as necessary.

For example, in balancing the reaction between potassium carbonate (K₂CO₃) and zinc chloride (ZnCl₂), you adjust coefficients so that the number of atoms for potassium, carbon, oxygen, zinc, and chlorine are the same on both sides of the equation: \[\mathrm{K}_{2}\mathrm{CO}_{3}(\mathrm{aq}) + \mathrm{ZnCl}_{2}(\mathrm{aq}) \rightarrow \mathrm{ZnCO}_{3}(\mathrm{s}) + 2\mathrm{KCl}(\mathrm{aq})\]Ensuring equations are balanced is essential for accurately predicting the amounts of reactants and products involved in a reaction.

Net Ionic Equation

A net ionic equation simplifies a chemical equation by showing only the ions involved in forming the precipitate. This excludes spectator ions that do not participate in the reaction. It provides a clearer view of the actual chemical change occurring.
To write a net ionic equation, follow these guidelines:

• Write the balanced molecular equation including states of matter.
• Separate all soluble ionic compounds into their ions.
• Identify and cancel out spectator ions (ions that appear unchanged on both sides).
• Write the remaining ions and compounds, showing the formation of the precipitate.

For example, when zinc ions (Zn²⁺) react with carbonate ions (CO₃²⁻), they form a solid precipitate of zinc carbonate (ZnCO₃). The net ionic equation only includes these components: \[\mathrm{Zn}^{2+}(\mathrm{aq}) + \mathrm{CO}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{ZnCO}_{3}(\mathrm{s})\]This simplified equation highlights the essence of the reaction by focusing solely on the formation of the precipitate, aiding in understanding the actual chemical change involved.