Problem 100 - Chemistry and Chemical Reactivity

Question

Balance these reactions and then classify each one as a precipitation, acid- base, or gas-forming reaction. Show states for the products \((\mathrm{s}, \ell, \mathrm{g}, \mathrm{aq}),\) and write the net ionic equation. (a) \(\mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{aq})+\mathrm{HClO}_{4}(\mathrm{aq}) \longrightarrow \mathrm{KClO}_{4}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{FeCl}_{2}(\mathrm{aq})+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}(\mathrm{aq}) \longrightarrow \mathrm{FeS}+\mathrm{NH}_{4} \mathrm{Cl}\) (c) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \longrightarrow \mathrm{FeCO}_{3}+\mathrm{NaNO}_{3}\)

Step-by-Step Solution

Answer

(a) Acid-base: \(\mathrm{CO}_{3}^{2-} + 2\mathrm{H}^{+} \to \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\); (b) Precipitation: \(\mathrm{Fe}^{2+} + \mathrm{S}^{2-} \to \mathrm{FeS}\); (c) Precipitation: \(\mathrm{Fe}^{2+} + \mathrm{CO}_{3}^{2-} \to \mathrm{FeCO}_{3}\).

Step 1: Balance Reaction (a)

Start by balancing the chemical equation: \[ \mathrm{K}_{2}\mathrm{CO}_{3}(\mathrm{aq}) + 2 \mathrm{HClO}_{4}(\mathrm{aq}) \longrightarrow 2 \mathrm{KClO}_{4}(\mathrm{aq}) + \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2}\mathrm{O}(\ell) \]Now, it is balanced with 2 molecules of \(\mathrm{HClO}_4\) and 2 molecules of \(\mathrm{KClO}_4\).

Step 2: Classify Reaction (a)

This reaction is an acid-base reaction because it involves the interaction between a carbonate (base) and an acid (\(\mathrm{HClO}_4\)), resulting in the formation of water and carbon dioxide.

Step 3: Net Ionic Equation for Reaction (a)

The net ionic equation is: \[ \mathrm{CO}_{3}^{2-}(\mathrm{aq}) + 2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2}\mathrm{O}(\ell) \]Spectator ions \(\mathrm{ClO}_{4}^{-}\) and \(\mathrm{K}^+\) are omitted from the net ionic equation.

Step 4: Balance Reaction (b)

Start by balancing the chemical equation: \[ \mathrm{FeCl}_{2}(\mathrm{aq}) + \left(\mathrm{NH}_{4}\right)_{2}\mathrm{S}(\mathrm{aq}) \longrightarrow \mathrm{FeS}(\mathrm{s}) + 2 \mathrm{NH}_{4}\mathrm{Cl}(\mathrm{aq}) \]The equation is balanced.

Step 5: Classify Reaction (b)

This is a precipitation reaction because \(\mathrm{FeS}\), a solid, is formed from the reaction in aqueous solution.

Step 6: Net Ionic Equation for Reaction (b)

The net ionic equation for reaction (b) is: \[ \mathrm{Fe}^{2+}(\mathrm{aq}) + \mathrm{S}^{2-}(\mathrm{aq}) \longrightarrow \mathrm{FeS}(\mathrm{s}) \]Spectator ions \(\mathrm{Cl}^-\) and \(\mathrm{NH}_{4}^{+}\) are omitted.

Step 7: Balance Reaction (c)

Balance the chemical equation: \[ \mathrm{Fe(NO}_{3})_{2}(\mathrm{aq}) + \mathrm{Na}_{2}\mathrm{CO}_{3}(\mathrm{aq}) \longrightarrow \mathrm{FeCO}_{3}(\mathrm{s}) + 2 \mathrm{NaNO}_{3}(\mathrm{aq}) \]The equation is balanced.

Step 8: Classify Reaction (c)

This is a precipitation reaction because \(\mathrm{FeCO}_{3}\), a solid, is formed from the reaction in aqueous solution.

Step 9: Net Ionic Equation for Reaction (c)

The net ionic equation for reaction (c) is: \[ \mathrm{Fe}^{2+}(\mathrm{aq}) + \mathrm{CO}_{3}^{2-}(\mathrm{aq}) \longrightarrow \mathrm{FeCO}_{3}(\mathrm{s}) \]Spectator ions \(\mathrm{Na}^+\) and \(\mathrm{NO}_3^-\) are omitted.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations ensures that the same number of each type of atom appears on both sides of the equation. This is vital for adhering to the Law of Conservation of Mass. To balance a chemical equation, the coefficients in front of each compound are adjusted. This process changes the quantity of molecules or formula units but keeps the actual substances unchanged.
Consider the example of reaction (a):

• The initial equation: \(\mathrm{K}_{2}\mathrm{CO}_{3}(\mathrm{aq}) + \mathrm{HClO}_{4}(\mathrm{aq}) \longrightarrow \mathrm{KClO}_{4} + \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\).
• By adjusting the coefficients to balance the equation, it becomes \(\mathrm{K}_{2}\mathrm{CO}_{3}(\mathrm{aq}) + 2 \mathrm{HClO}_{4}(\mathrm{aq}) \longrightarrow 2 \mathrm{KClO}_{4}(\mathrm{aq}) + \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2}\mathrm{O}(\ell)\).
• This ensures all atoms on the reactant side match those on the product side.

Balancing equations is a step-by-step process that involves counting atoms of each element carefully and adjusting the coefficients as needed. It can sometimes require trial and error, especially with complex equations.

Net Ionic Equations

Net ionic equations show only the species involved in the actual chemical change occurring in a reaction, leaving out spectator ions that do not participate in the reaction. These equations highlight the primary reaction events.
For reaction (a), the net ionic equation is:

• \(\mathrm{CO}_{3}^{2-}(\mathrm{aq}) + 2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2}\mathrm{O}(\ell)\).
• Spectator ions \(\mathrm{ClO}_{4}^{-}\) and \(\mathrm{K}^+\) are omitted.

Writing the net ionic equation involves the following steps:

• Write the balanced molecular equation.
• Dissociate all strong electrolytes into their respective ions.
• Identify and omit the spectator ions.
• Simplify to the net ionic form.

This approach emphasizes the chemical species that are actively engaged in the reaction.

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions combine to form a solid or precipitate. This process is essential for understanding solubility rules and how they govern reaction outcomes.
A clear example is reaction (b):

• \(\mathrm{FeCl}_{2}(\mathrm{aq}) + (\mathrm{NH}_{4})_{2}\mathrm{S}(\mathrm{aq}) \longrightarrow \mathrm{FeS}(\mathrm{s}) + 2 \mathrm{NH}_{4}\mathrm{Cl}(\mathrm{aq})\).
• The formation of \(\mathrm{FeS}(\mathrm{s})\) signifies the precipitation reaction.

Precipitation reactions are generally predictable through use of solubility rules:

• Most sulfides, carbonates, and hydroxides are insoluble.
• Exceptions to these rules are common for alkali metals and ammonium compounds.

Understanding these reactions is critical in fields like chemistry and environmental science where removing impurities is necessary.

Acid-Base Reactions

Acid-base reactions involve the transfer of protons between reactants. These reactions are often represented as either neutralization or proton exchange reactions and are key in maintaining pH balance in various systems.
In reaction (a), the interaction between \(\mathrm{K}_{2}\mathrm{CO}_{3}\) and \(\mathrm{HClO}_{4}\) is classified as an acid-base reaction:

• Here, \(\mathrm{HClO}_{4}\) (acid) donates protons to the carbonate ion \(\mathrm{CO}_{3}^{2-}\) (base).
• The result is carbon dioxide and water, typical products of an acid-base reaction.

These reactions can be further categorized:

• Neutralization reactions: Often produce water and a salt.
• Proton transfer reactions: Involve acids and bases that transfer protons to create conjugate forms.

Mastery of acid-base reactions is essential for studying chemical equilibria and biological systems.

Gas-Forming Reactions

Gas-forming reactions are characterized by the evolution of a gaseous product. These reactions are exciting because they often produce bubbling or fizzing, visible signs of gas formation. They are a subset of decompositions or acid-base reactions.
For instance, consider reaction (a):

• \(\mathrm{K}_{2}\mathrm{CO}_{3}(\mathrm{aq}) + 2 \mathrm{HClO}_{4}(\mathrm{aq}) \longrightarrow 2 \mathrm{KClO}_{4}(\mathrm{aq}) + \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2}\mathrm{O}(\ell)\)
• This reaction releases \(\mathrm{CO}_{2}(\mathrm{g})\), indicating a gas-forming reaction.

Common gas-forming reactions include:

• Decomposition of carbonates or bicarbonates when reacted with acids, releasing \(\mathrm{CO}_{2}\).
• Reactions between acids and sulfides, releasing \(\mathrm{H}_{2}\mathrm{S}\).

Recognizing these reactions helps in analytical chemistry and industrial processes where gas release or capture is significant.