Problem 73 - Chemistry and Chemical Reactivity

Question

Balance equations for these reactions that occur in aqueous solution, and then classify each as a precipitation, acid-base, or gas-forming reaction. Show states for the products \((\mathrm{s}, \ell, \mathrm{g}, \mathrm{aq}),\) give their names, and write the net ionic equation. (a) \(\mathrm{K}_{2} \mathrm{CO}_{3}+\mathrm{HClO}_{4} \rightarrow \mathrm{KClO}_{4}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{FeCl}_{2}+\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S} \rightarrow \mathrm{FeS}+\mathrm{NH}_{4} \mathrm{Cl}\) (c) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \rightarrow \mathrm{FeCO}_{3}+\mathrm{NaNO}_{3}\) (d) \(\mathrm{NaOH}+\mathrm{FeCl}_{3} \rightarrow \mathrm{NaCl}+\mathrm{Fe}(\mathrm{OH})_{3}\)

Step-by-Step Solution

Answer

(a) Gas-forming; (b)-(d) Precipitation reactions. Net ionic: (a) \(\mathrm{CO}_{3}^{2-} + 2\mathrm{H}^{+} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O}\); (b) \(\mathrm{Fe}^{2+} + \mathrm{S}^{2-} \rightarrow \mathrm{FeS}\); (c) \(\mathrm{Fe}^{2+} + \mathrm{CO}_{3}^{2-} \rightarrow \mathrm{FeCO}_{3}\); (d) \(\mathrm{Fe}^{3+} + 3\mathrm{OH}^{-} \rightarrow \mathrm{Fe}(\mathrm{OH})_{3}\).

Step 1: Balance the Chemical Equations

We begin by balancing each chemical equation provided, ensuring that the number of atoms for each element is the same on both sides of the equation.(a) \(\mathrm{K}_{2} \mathrm{CO}_{3} + 2\mathrm{HClO}_{4} \rightarrow 2\mathrm{KClO}_{4} + \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O}\)(b) \(\mathrm{FeCl}_{2} + \left(\mathrm{NH}_{4}\right)_{2} \mathrm{S} \rightarrow \mathrm{FeS} + 2\mathrm{NH}_{4} \mathrm{Cl}\)(c) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2} + \mathrm{Na}_{2} \mathrm{CO}_{3} \rightarrow \mathrm{FeCO}_{3} + 2\mathrm{NaNO}_{3}\)(d) \(3\mathrm{NaOH} + \mathrm{FeCl}_{3} \rightarrow 3\mathrm{NaCl} + \mathrm{Fe} (\mathrm{OH})_{3}\)

Step 2: Determine the State of Each Compound

Next, we identify the state (solid, liquid, gas, aqueous) of each compound in the reactions.(a) \(\mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{aq}) + 2\mathrm{HClO}_{4}(\mathrm{aq}) \rightarrow 2\mathrm{KClO}_{4}(\mathrm{aq}) + \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\ell)\)(b) \(\mathrm{FeCl}_{2}(\mathrm{aq}) + \left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}(\mathrm{aq}) \rightarrow \mathrm{FeS}(\mathrm{s}) + 2\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})\)(c) \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) + \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \rightarrow \mathrm{FeCO}_{3}(\mathrm{s}) + 2\mathrm{NaNO}_{3}(\mathrm{aq})\)(d) \(3\mathrm{NaOH}(\mathrm{aq}) + \mathrm{FeCl}_{3}(\mathrm{aq}) \rightarrow 3\mathrm{NaCl}(\mathrm{aq}) + \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\)

Step 3: Classify Each Reaction

Identify the type of reaction based on the products formed:(a) **Gas-forming reaction**: The formation of \(\mathrm{CO}_{2}(\mathrm{g})\) classifies this as a gas-forming reaction.(b) **Precipitation reaction**: \(\mathrm{FeS}(\mathrm{s})\) is formed as a precipitate.(c) **Precipitation reaction**: \(\mathrm{FeCO}_{3}(\mathrm{s})\) precipitates from solution.(d) **Precipitation reaction**: The insoluble \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) is formed.

Step 4: Write the Net Ionic Equations

Extract the net ionic equation by removing spectator ions.(a) \(\mathrm{CO}_{3}^{2-}(\mathrm{aq}) + 2\mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\ell)\)(b) \(\mathrm{Fe}^{2+}(\mathrm{aq}) + \mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{FeS}(\mathrm{s})\)(c) \(\mathrm{Fe}^{2+}(\mathrm{aq}) + \mathrm{CO}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{FeCO}_{3}(\mathrm{s})\)(d) \(\mathrm{Fe}^{3+}(\mathrm{aq}) + 3\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\)

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a crucial step when working with reactions. It ensures that the law of conservation of mass is obeyed. This law states that matter is neither created nor destroyed during a chemical reaction. This means the number of atoms for each element must be equal on both sides of the chemical equation. To balance an equation, adjust the coefficients, which are the numbers before the compounds.
Select appropriate coefficients that will make the number of atoms of each element equal on both sides. For instance, in the reaction \(\mathrm{K}_{2} \mathrm{CO}_{3} + 2\mathrm{HClO}_{4} \rightarrow 2\mathrm{KClO}_{4} + \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O}\), noticing that there are 2 potassium atoms and balancing them across the reactants and products is crucial. This way, both sides have the same number of each type of atom, satisfying the conservation of mass.

Types of Chemical Reactions

Chemical reactions can be classified into different types based on the nature of the reactants and products. Each type reveals specific processes taking place during the reaction.

• Gas-forming reactions: These involve the release of a gaseous product. For example, when mixing \(\mathrm{K}_{2} \mathrm{CO}_{3} \) with \(\mathrm{HClO}_{4}\), \(\mathrm{CO}_{2}\) forms as gas.
• Precipitation reactions: These occur when two aqueous solutions combine to form a solid precipitate. A good example is the formation of \(\mathrm{FeS}\) in the reaction between \(\mathrm{FeCl}_{2}\) and \((\mathrm{NH}_{4})_{2} \mathrm{S}\).
• Acid-base reactions: Involve the exchange of ions between an acid and a base. While not directly shown, these often result in water formation.

Understanding the reaction type helps in predicting the behavior of the reactants and the products formed.

Net Ionic Equations

Net ionic equations provide a simplified look at a chemical reaction by highlighting only the particles that undergo a change, omitting the spectator ions. Spectator ions are ions that appear on both sides of the complete ionic equation and do not participate in the reaction. Let's see some examples:

• In the reaction with \(\mathrm{K}_{2} \mathrm{CO}_{3}\) and \(\mathrm{HClO}_{4}\), the net ionic equation is \[\mathrm{CO}_{3}^{2-}(\mathrm{aq}) + 2\mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\ell)\]. Only the carbonate and hydrogen ions react here.
• For the precipitation reaction, \(\mathrm{Fe}^{2+}(\mathrm{aq}) + \mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{FeS}(\mathrm{s})\), only the ions forming the solid \(\mathrm{FeS}\) are shown.

By understanding and writing net ionic equations, you can focus on the key changes occurring during the reaction.

States of Matter in Chemical Reactions

When writing chemical equations, it’s important to indicate the states of matter of the reactants and products: solids (s), liquids (l), gases (g), and aqueous solutions (aq). These states provide essential context:

• Solids, such as \(\mathrm{FeS}(\mathrm{s})\), indicate precipitates formed during a reaction.
• Liquids are pure substances that are in a distinct liquid state, such as water \(\mathrm{H}_{2} \mathrm{O}(\ell)\).
• Gases include volatile compounds released or used in a reaction, like \(\mathrm{CO}_{2}(\mathrm{g})\).
• Aqueous solutions, denoted as \(\mathrm{aq}\), are substances dissolved in water, allowing ions to move freely. For instance, \(\mathrm{KClO}_{4}(\mathrm{aq})\) exists in solution.

Indicating the states of matter ensures a clear understanding of reaction conditions and the phases in which various substances exist.