Problem 49 - General Chemistry: Principles and Modern Applications

Question

Write plausible chemical equations for the (a) dissolving of lead(II) oxide in nitric acid; (b) heating of \(\operatorname{snCO}_{3}(\mathrm{s}) ;\) (c) reduction of lead(II) oxide by carbon; (d) reduction of \(\mathrm{Fe}^{3+}(\mathrm{aq})\) to \(\mathrm{Fe}^{2+}(\mathrm{aq})\) by \(\mathrm{Sn}^{2+}(\mathrm{aq});\) (e) formation of lead(II) sulfate during high-temperature roasting of lead(II) sulfide.

Step-by-Step Solution

Answer

(a) \(\mathrm{PbO (s) + 2HNO}_{3}(\mathrm{l}) \rightarrow \mathrm{Pb(NO}_{3})_{2}(\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O (l)}\), (b) \(\mathrm{SnCO}_{3}(\mathrm{s}) \rightarrow \mathrm{SnO (s) + CO}_{2}(\mathrm{g})\), (c) \(\mathrm{PbO (s) + C (s) \rightarrow Pb (s) + CO}_{2}(\mathrm{g})\), (d) \(\mathrm{2Fe}^{3+}(\mathrm{aq}) + \(\mathrm{Sn}^{2+}(\mathrm{aq}) \rightarrow \mathrm{2Fe}^{2+}(\mathrm{aq}) + \mathrm{Sn}^{4+}(\mathrm{aq})\), (e) \(\mathrm{2PbS (s) + 3O}_{2}(\mathrm{g}) \rightarrow \mathrm{2PbSO}_{4}(\mathrm{s}) + \mathrm{2SO}_{2}(\mathrm{g})\).

Step 1: Dissolving of lead(II) oxide in nitric acid

The dissolving of a metal oxide in an acid typically results in the formation of a salt and water. In this case, lead(II) oxide (\(\mathrm{PbO}\)) reacts with nitric acid (\(\mathrm{HNO}_{3}\)) to produce lead(II) nitrate (\(\mathrm{Pb(NO}_{3})_{2}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)). Therefore, the balanced chemical equation is: \(\mathrm{PbO (s) + 2HNO}_{3}(\mathrm{l}) \rightarrow \mathrm{Pb(NO}_{3})_{2}(\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O (l)}\).

Step 2: Heating of \(\mathrm{SnCO}_{3}(\mathrm{s})\)

When heated, metal carbonates decompose to produce a metal oxide and carbon dioxide. Thus, \(\mathrm{SnCO}_{3}(\mathrm{s})\) yields \(\mathrm{SnO(s)}\) and \(\mathrm{CO}_{2}(\mathrm{g})\). The balanced equation is: \(\mathrm{SnCO}_{3}(\mathrm{s}) \rightarrow \mathrm{SnO (s) + CO}_{2}(\mathrm{g})\).

Step 3: Reduction of lead(II) oxide by carbon

The reduction of a metal oxide using carbon involves the conversion of the metal to its elemental form and the formation of carbon dioxide. Hence, \(\mathrm{PbO(s)}\) reacts with \(\mathrm{C(s)}\) to give \(\mathrm{Pb(s)}\) and \(\mathrm{CO}_{2}(\mathrm{g})\). The balanced chemical equation is: \(\mathrm{PbO (s) + C (s) \rightarrow Pb (s) + CO}_{2}(\mathrm{g})\).

Step 4: Reduction of \(\mathrm{Fe}^{3+}(\mathrm{aq})\) to \(\mathrm{Fe}^{2+}(\mathrm{aq})\) by \(\mathrm{Sn}^{2+}(\mathrm{aq})\)

This is a redox reaction where \(\mathrm{Sn}^{2+}\) gets oxidized to \(\mathrm{Sn}^{4+}\) and \(\mathrm{Fe}^{3+}\) gets reduced to \(\mathrm{Fe}^{2+}\). The balanced chemical reaction is: \(\mathrm{2Fe}^{3+}(\mathrm{aq}) + \mathrm{Sn}^{2+}(\mathrm{aq}) \rightarrow \mathrm{2Fe}^{2+}(\mathrm{aq}) + \mathrm{Sn}^{4+}(\mathrm{aq})\).

Step 5: Formation of lead(II) sulfate during high-temperature roasting of lead(II) sulfide

Upon roasting, lead(II) sulfide reacts with oxygen present in air to generate lead(II) sulfate and sulfur dioxide. The balanced chemical equation is: \(\mathrm{2PbS (s) + 3O}_{2}(\mathrm{g}) \rightarrow \mathrm{2PbSO}_{4}(\mathrm{s}) + \mathrm{2SO}_{2}(\mathrm{g})\).

Key Concepts

Metal Oxide Reactions

Metal oxides play a significant role in various chemical reactions, particularly in interactions with acids and their role in reduction processes. Understanding these reactions helps us unlock a wide range of chemical applications, from industrial processing to environmental chemistry.

Metal oxides typically react with acids to form salt and water in a process known as a neutralization reaction. The general form of this reaction can be represented as: \( \text{Metal Oxide} + \text{Acid} \rightarrow \text{Salt} + \text{Water} \). In the exercise provided, the reaction between lead(II) oxide (a metal oxide) and nitric acid (an acid) results in a specific type of salt known as lead(II) nitrate and water as byproducts.

This type of reaction can be applied to various metal oxides and acids, leading to the creation of a broad range of salts. When working with these reactions, it's important to balance the equation to adhere to the law of conservation of mass, ensuring the same number of atoms of each element on both sides of the equation.

Thermal Decomposition of Carbonates

Thermal decomposition is a chemical process where a compound breaks down when heated. Metal carbonates, in particular, tend to decompose into metal oxides and carbon dioxide gas when exposed to high temperatures, an important process in materials science and geochemistry.

The general equation for the thermal decomposition of a metal carbonate is given by: \( \text{Metal Carbonate} \rightarrow \text{Metal Oxide} + \text{Carbon Dioxide} \). The exercise example with tin carbonate (\( \text{SnCO}_3 \)) breaking down into tin oxide (\( \text{SnO} \)) and carbon dioxide (\( \text{CO}_2 \)) shows this type of reaction.

This understanding is critical when processing ores or manufacturing materials where precise control of the components is required. It's also a tangible illustration of how heating can induce chemical changes, reinforcing concepts such as endothermic reactions and chemical stability.

Redox Reactions

Redox reactions are a fundamental type of chemical reaction that involves the transfer of electrons between reactants, categorizing them into two processes: reduction and oxidation. These reactions are pivotal in energy generation, metallurgy, and biological systems.

In a redox process, the substance that loses electrons is 'oxidized,' while the one that gains electrons is 'reduced.' The reaction provided in the exercise, where \( \text{Fe}^{3+} \) is reduced to \( \text{Fe}^{2+} \) by \( \text{Sn}^{2+} \), showcases a classic redox reaction where tin is oxidized and iron is reduced.

To recognize these reactions, remember the mnemonic 'OIL RIG' - Oxidation Is Loss, Reduction Is Gain - referring to the loss or gain of electrons, respectively. Redox reactions are integral to countless processes, including the function of batteries, biogeochemical cycles, and the production of metals.

Roasting of Sulfides

Roasting is a metallurgical process involving the heating of sulfide ores in the presence of air. It's primarily used to remove sulfur from the ore and to convert the sulfides into oxides, preparing them for further refinement, such as smelting.

During roasting, a chemical reaction occurs where the metal sulfide is transformed into a metal oxide and sulfur dioxide gas is released. A general equation for this reaction is: \( \text{Metal Sulfide} + \text{Oxygen} \rightarrow \text{Metal Oxide} + \text{Sulfur Dioxide} \). The lead(II) sulfide example from the exercise, turning into lead(II) sulfate, exemplifies this process. Additionally, it's important to manage the environmental impacts of releasing sulfur dioxide, a compound known for contributing to acid rain.

Understanding these reactions not only aids in metal extraction but also in developing methods to minimize the environmental footprint of mining activities.