Question

Complete and balance each of the following equations: (a) \(\mathrm{Cr}_{2} \mathrm{O}_{3}(s)+\mathrm{Na}(l) \longrightarrow\) (b) \(\mathrm{PbCO}_{3}(s) \stackrel{\Delta}{\longrightarrow}\) (c) \(\mathrm{CdS}(s)+\mathrm{O}_{2}(g) \stackrel{\Delta}{\longrightarrow}\) (d) \(\mathrm{ZnO}(s)+\mathrm{CO}(g) \stackrel{\Delta}{\longrightarrow}\)

Short answer

The short answers are: a) \(6 Na(l) + Cr_{2}O_{3}(s) \longrightarrow 3 Na_{2}O(s) + 2 Cr(s)\) b) \(PbCO_{3}(s) \stackrel{\Delta}{\longrightarrow} PbO(s) + CO_{2}(g)\) c) \(2 CdS(s) + 3 O_{2}(g) \stackrel{\Delta}{\longrightarrow} 2 CdO(s) + 2 SO_{2}(g)\) d) \(ZnO(s) + CO(g) \stackrel{\Delta}{\longrightarrow} Zn(s) + CO_{2}(g)\)

Step-by-step solution

(a) Step 1: Find the products of the reaction

The given reactants are Cr2O3 (Chromium(III) oxide) and Na (Sodium). In this type of reaction, Sodium will reduce the chromium in chromium(III) oxide, and will itself be oxidized, leading to the formation of Sodium metal and chromium metal. The products are Na2O (Sodium oxide) and Cr (Chromium). The unbalanced equation can be written as: Cr2O3(s) + Na(l) -> Na2O(s) + Cr(s)

(a) Step 2: Balance the equation

To balance the equation, we should have the same number of each kind of atom on both sides. We can adjust the coefficients in front of the reactants and products: \(6\: Na(l) + Cr_{2}O_{3}(s) -> 3 Na_{2}O(s) + 2 Cr(s)\) The balanced equation for the reaction (a) is: \(6 Na(l) + Cr_{2}O_{3}(s) \longrightarrow 3 Na_{2}O(s) + 2 Cr(s)\)

(b) Step 1: Thermal decomposition of PbCO3

PbCO3 (Lead(II) carbonate) will decompose into its individual components under heat. Based on the type of compound, the products of the decomposition will be PbO (Lead(II) oxide) and CO2 (Carbon dioxide). The unbalanced equation can be written as: PbCO3(s) -> PbO(s) + CO2(g)

(b) Step 2: Balance the equation

In this case, the equation is already balanced, as there is the same number of each kind of atom on both sides. The balanced equation for the reaction (b) is: \(PbCO_{3}(s) \stackrel{\Delta}{\longrightarrow} PbO(s) + CO_{2}(g)\)

(c) Step 1: Reaction of CdS and O2

The given reactants are CdS (Cadmium Sulfide) and O2 (Oxygen). This is a redox reaction in which CdS will get oxidized. The expected products would be CdO (Cadmium oxide) and SO2 (Sulfur dioxide). The unbalanced equation can be written as: CdS(s) + O2(g) -> CdO(s) + SO2(g)

(c) Step 2: Balance the equation

To balance the equation, we adjust the coefficients in front of the reactants and products: 2 CdS(s) + 3 O2(g) -> 2 CdO(s) + 2 SO2(g) The balanced equation for the reaction (c) is: \(2 CdS(s) + 3 O_{2}(g) \stackrel{\Delta}{\longrightarrow} 2 CdO(s) + 2 SO_{2}(g)\)

(d) Step 1: Reaction of ZnO and CO

The given reactants are ZnO (Zinc oxide) and CO (Carbon monoxide). In this reaction, CO will reduce Zn in ZnO resulting in the formation of Zn (Zinc) and CO2 (Carbon dioxide). The unbalanced equation can be written as: ZnO(s) + CO(g) -> Zn(s) + CO2(g)

(d) Step 2: Balance the equation

In this case, the equation is already balanced, as there is the same number of each kind of atom on both sides. The balanced equation for the reaction (d) is: \(ZnO(s) + CO(g) \stackrel{\Delta}{\longrightarrow} Zn(s) + CO_{2}(g)\)

Key concepts

Balancing Equations

Balancing chemical equations is pivotal in understanding chemical reactions. Essentially, it's about ensuring the same number of each type of atom appears on both sides of a reaction. This mirrors real-world observations, where matter cannot be created or destroyed. A balanced equation tells us the precise proportions of reactants and products involved in a reaction.
For example, in reaction (a) with Chromium(III) oxide and sodium, the initial equation may seem incomplete. Simply adding coefficients in front of each compound, like \[ 6 \text{Na}(l) + \text{Cr}_2\text{O}_3 \rightarrow 3 \text{Na}_2\text{O}(s) + 2 \text{Cr}(s) \] helps us obey the law of conservation of mass. This means no atoms are lost or gained in the reaction, they are merely rearranged. Using coefficients effectively can mean the difference between correctly predicting how a reaction takes place or not.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are essential in both biological and industrial processes. In these reactions, oxidation and reduction occur simultaneously: one substance loses electrons (oxidized), while another gains electrons (reduced). Take, for instance, reaction (c) involving cadmium sulfide and oxygen:\[ 2 \text{CdS}(s) + 3 \text{O}_2(g) \rightarrow 2 \text{CdO}(s) + 2 \text{SO}_2(g) \]Cadmium sulfide is oxidized, losing electrons and forming cadmium oxide, while oxygen is reduced to form sulfur dioxide.
Understanding redox reactions allows us to better grasp processes as simple as rusting or as complex as cellular respiration. If we can follow the movement of electrons during these reactions, we deepen our understanding of many natural and industrial phenomena.

Thermal Decomposition

Thermal decomposition is an intriguing concept wherein a compound breaks down into simpler substances when heated. It's a fascinating process, seen in reaction (b) when lead(II) carbonate decomposes:\[ \text{PbCO}_3(s) \rightarrow \text{PbO}(s) + \text{CO}_2(g) \]The compound absorbs heat, causing its chemical bonds to break and yield lead(II) oxide and carbon dioxide.
This type of reaction is crucial in various industries. For instance, in creating lime from limestone, or processing metals. The comprehension of these reactions gives insight into the stability of compounds and energy changes occurring during reactions. Recognizing the signs of thermal decomposition, such as gas evolution and color change, is valuable in both academic and practical contexts.

Oxidation-Reduction Processes

Oxidation-reduction (redox) processes are fundamental in chemistry and stoichiometry. They involve the transfer of electrons between substances, driving many chemical reactions. In the given exercise, we see this vividly in cases like the reduction of zinc oxide:\[ \text{ZnO}(s) + \text{CO}(g) \rightarrow \text{Zn}(s) + \text{CO}_2(g) \]Here, carbon monoxide takes up oxygen, reducing zinc oxide to zinc and oxidizing itself to carbon dioxide. The process underscores the principle that the total transfer of electrons must be balanced, ensuring charge conservation.
Grasping these processes fortifies our comprehension of everything from metabolism in living organisms to the operation of electrochemical cells. Familiarizing ourselves with oxidizing and reducing agents is key, as it allows us to predict and balance redox reactions with greater confidence. Understanding the broader scope of these processes can enhance our foundational knowledge of chemical transformations.