Problem 23 - Chemistry and Chemical Reactivity

Question

Balance the following equations, and then write the net ionic equation. (a) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(\mathrm{aq})+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \longrightarrow\) \(\mathrm{CuCO}_{3}(\mathrm{s})+\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{aq})\) (b) \(\mathrm{Pb}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{PbCl}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (c) \(\mathrm{BaCO}_{3}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow\) \(\mathrm{BaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g})\)

Step-by-Step Solution

Answer

Net ionic equations are: (a) \(\mathrm{Cu}^{2+} + \mathrm{CO}_{3}^{2-} \rightarrow \mathrm{CuCO}_{3}\); (b) same as balanced; (c) \(\mathrm{BaCO}_{3} + 2\mathrm{H}^{+} \rightarrow \mathrm{Ba}^{2+} + \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\).

Step 1: Balance Equation (a)

The given equation is \[ \left(\mathrm{NH}_{4}\right)_{2}\mathrm{CO}_{3}(\mathrm{aq})+\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) \rightarrow \mathrm{CuCO}_{3}(\mathrm{s})+\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{aq}) \]. We notice that all elements are already balanced. Thus, the balanced equation is the same as the given one.

Step 2: Write Net Ionic Equation for (a)

To write the net ionic equation, we first recognize the key change, which is the formation of \(\mathrm{CuCO}_{3}(\mathrm{s})\) from \(\mathrm{Cu}^{2+}(\mathrm{aq})\) and \(\mathrm{CO}_{3}^{2-}(\mathrm{aq})\). The net ionic equation is \[ \mathrm{Cu}^{2+}(\mathrm{aq}) + \mathrm{CO}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{CuCO}_{3}(\mathrm{s}) \].

Step 3: Balance Equation (b)

The unbalanced equation is \[ \mathrm{Pb}(\mathrm{OH})_{2}(\mathrm{s}) + \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{PbCl}_{2}(\mathrm{s}) + \mathrm{H}_{2}\mathrm{O}(\ell) \]. We see that 2 moles of \(\mathrm{HCl}\) are needed to react with 1 mole of \(\mathrm{Pb}(\mathrm{OH})_{2}\). Thus, the balanced equation is \[ \mathrm{Pb}(\mathrm{OH})_{2}(\mathrm{s}) + 2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{PbCl}_{2}(\mathrm{s}) + 2 \mathrm{H}_{2}\mathrm{O}(\ell) \].

Step 4: Write Net Ionic Equation for (b)

Since \(\mathrm{PbCl}_{2}(\mathrm{s})\) is a solid, the net ionic equation is essentially the same as the balanced equation, except we don't break down the solid. Thus, no need to write ions as it involves all solids and unchanged species.

Step 5: Balance Equation (c)

The given equation is \[ \mathrm{BaCO}_{3}(\mathrm{s}) + \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{BaCl}_{2}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\ell) + \mathrm{CO}_{2}(\mathrm{g}) \]. Balancing it, we find that we need 2 moles of \(\mathrm{HCl}\) per mole of \(\mathrm{BaCO}_{3}\). The balanced equation is \[ \mathrm{BaCO}_{3}(\mathrm{s}) + 2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{BaCl}_{2}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\ell) + \mathrm{CO}_{2}(\mathrm{g}) \].

Step 6: Write Net Ionic Equation for (c)

In solution, \(\mathrm{BaCl}_{2}(aq)\) dissociates into ions, and thus, the net ionic equation will focus on the direct reaction with the \(\mathrm{H}^+\) ions and carbonate. The net ionic equation is \[ \mathrm{BaCO}_{3}(\mathrm{s}) + 2\mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Ba}^{2+}(\mathrm{aq}) + \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2}\mathrm{O}(\ell) \].

Key Concepts

Net Ionic Equations: Simplifying Reactions

In chemistry, understanding a net ionic equation involves distilling a chemical reaction to its essence, showing only the particles directly involved in the change. When you write a net ionic equation, you focus solely on the ions or molecules that actually participate in the reaction. This simplifies the equation and clarifies what occurs at a molecular level.

To write a net ionic equation, follow these steps:

• Start with a balanced molecular equation, ensuring that the number of atoms for each element is equal on both sides.
• Dissociate all strong electrolytes into their ions in the aqueous state. For weak electrolytes or insoluble substances, keep them together.
• Identify and cancel out the spectator ions, which appear unchanged on both sides of the equation.
• Finalize by presenting only the particles that change, forming the net ionic equation.

For example, in exercise (a), \[\mathrm{Cu}^{{2+}}(\mathrm{aq}) + \mathrm{CO}_3^{2-}(\mathrm{aq}) \rightarrow \mathrm{CuCO}_3(\mathrm{s})\]the net ionic equation shows just the formation of solid copper carbonate, skipping the unreactive parts. This approach helps to highlight the key components of chemical interactions and emphasizes the transformation that takes place.

Understanding Chemical Reactions

Chemical reactions are transformations where substances, known as reactants, change into new substances, called products. These reactions can be either physical changes or chemical changes involving the breaking and forming of bonds.

Basic types of chemical reactions include:

• Synthesis reactions: Two or more simple substances combine to form a more complex product. Example: \[\mathrm{A} + \mathrm{B} \rightarrow \mathrm{AB}\]
• Decomposition reactions: A complex substance breaks down into simpler ones. Example: \[\mathrm{AB} \rightarrow \mathrm{A} + \mathrm{B}\]
• Single replacement reactions: An element in a compound is replaced by another element. Example:\[\mathrm{A} + \mathrm{BC} \rightarrow \mathrm{AC} + \mathrm{B}\]
• Double replacement reactions: The ions of two compounds exchange places. Example:\[\mathrm{AB} + \mathrm{CD} \rightarrow \mathrm{AD} + \mathrm{CB}\]
• Combustion reactions: A substance combines with oxygen, releasing energy. Example:\[\mathrm{C}_x\mathrm{H}_y + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}\]

Balancing these reactions is crucial, as it ensures the conservation of mass, meaning the same number of each type of atom is present in the reactants and products. Understanding these basic reaction types provides a foundation to explore more specific chemical processes, such as those involving the changes of state or electron transfer.

Mastering Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It is derived from the Greek words 'stoicheion' (element) and 'metron' (measure), depicting the measurement of elements involved in a chemical reaction.

Key parts of stoichiometry in a chemical equation include:

• Moles: The basic unit of measure in chemistry. One mole equals Avogadro's number, \(6.022 \times 10^{23}\), of particles.
• Mole ratios: Ratios derived from the coefficients of a balanced chemical equation that relate the amount of reactants to products.
• Limiting reactant: The reactant that is completely consumed in a reaction, determining the maximum amount of product that can be formed.
• Percent yield: Comparing the actual yield, the amount of product made, to the theoretical yield, the maximum possible amount of product, expressed as a percentage.

For example, in the reaction between hydrochloric acid and barium carbonate to form barium chloride, water and carbon dioxide, stoichiometry helps you predict how much product you will get given the initial amounts of reactants.
By applying stoichiometry, you ensure precision in chemical experiments and processes, enabling chemists to optimize conditions and materials used.