Step 1: Identification of Reactants and Products

Based on the provided chemical description, the reactants and their respective states are:

• Hydrogen gas ($H_{2(g)}$)
• Chlorine gas ($Cl_{2(g)}$)

The product formed is Hydrogen chloride gas ($HCl_{(g)}$).

Step 2: Skeletal Chemical Equation

The unbalanced chemical equation representing the transformation is:

$H_{2(g)} + Cl_{2(g)} \rightarrow HCl_{(g)}$

Step 3: Balancing the Chemical Equation

To satisfy the [Law of Conservation of Mass], which states that matter cannot be created or destroyed in a chemical reaction, the number of atoms of each element must be equal on both sides of the equation.

To balance the atoms, we apply a stoichiometric coefficient of $2$ to the product $HCl$ [Per the Trial and Error Method for Balancing Equations]:

$H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$

Step 4: Identification of Reaction Type

The reaction involves two simple substances combining to form a single, more complex substance. [By the definition of Combination Reaction ($A + B \rightarrow AB$)], this process is classified as a Combination Reaction.

Final Solution: The balanced chemical equation is $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$, and the reaction type is a Combination Reaction.

Step 1: Identification of Reactants and Reaction Type

The chemical reaction involves iron ($Fe$), which is a metal, and dilute hydrochloric acid ($HCl$). When a metal reacts with a dilute mineral acid, it typically undergoes a single displacement reaction, producing a metal salt and releasing hydrogen gas. [Per the Activity Series of Metals, iron is more reactive than hydrogen and can displace it from dilute acids.]

Step 2: Chemical Equation Formulation

The skeleton equation for the reaction is as follows:

$Fe(s) + HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)$

To ensure mass balance, we assign coefficients to satisfy the Law of Conservation of Mass:

$Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)$

In this reaction, iron displaces the hydrogen from the hydrochloric acid, forming iron(II) chloride (ferrous chloride) and releasing hydrogen gas as bubbles.

Step 3: Visual Representation of the Evolution of Gas

Step 4: Analytical Summary of Products

The interaction results in the formation of iron(II) chloride and the evolution of hydrogen gas. This is categorized as an exothermic displacement reaction.

Final Solution: Hydrogen gas and iron(II) chloride are produced.

Step 1: Chemical Formulae Identification

To construct the chemical equation, we translate the names of the reactants and products into their respective chemical formulas based on valence and ionic composition:

• Zinc carbonate: Zinc ($Zn^{2+}$) and Carbonate ($CO_3^{2-}$) form $ZnCO_3(s)$.
• Zinc oxide: Zinc ($Zn^{2+}$) and Oxide ($O^{2-}$) form $ZnO(s)$.
• Carbon dioxide: Carbon and Oxygen form $CO_2(g)$.

Step 2: Constructing the Skeletal Equation

The skeletal equation is represented as follows:

$ZnCO_3(s) \rightarrow ZnO(s) + CO_2(g)$

Step 3: Verification of Law of Conservation of Mass

Per the [Law of Conservation of Mass], the number of atoms of each element must be equal on both the reactant and product sides. We create an atom count table:

As the atom counts match, the equation is inherently balanced.

Step 4: Classification of Reaction Type

The reaction involves a single reactant breaking down into two or more simpler products upon the application of thermal energy (implied for carbonate decomposition). [Per the definition of Decomposition Reactions], this is categorized as a Thermal Decomposition Reaction.

Step 5: Theoretical Justification

The reaction $ZnCO_3(s) \xrightarrow{\Delta} ZnO(s) + CO_2(g)$ requires energy input to break the strong ionic bonds within the crystal lattice of the carbonate, resulting in the evolution of carbon dioxide gas [Per the Arrhenius concept of thermal decomposition].

Final Solution: The balanced equation is $ZnCO_3(s) \rightarrow ZnO(s) + CO_2(g)$; this is a Decomposition Reaction.

Step 1: Analyzing the Reactants and Products

The given chemical equation is: $Fe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe$.

To identify the type of reaction, we evaluate the displacement of elements:

• Reactants: Iron(III) oxide ($Fe_2O_3$) and Aluminium ($Al$).
• Products: Aluminium oxide ($Al_2O_3$) and Iron ($Fe$).

Observation: The Aluminium atom has displaced the Iron atom from its oxide compound. [By the Reactivity Series of Metals, Aluminium is more reactive than Iron and can therefore displace it from its salt solution or oxide.]

Step 2: Classifying the Reaction

The reaction follows the general form of a single displacement reaction: $A + BC \rightarrow AC + B$. Since one element (Al) replaces another element (Fe) from its compound ($Fe_2O_3$), this is classified as a Displacement Reaction.

Furthermore, evaluating the oxidation states:

Because Fe is reduced (gain of electrons) and Al is oxidized (loss of electrons), this is also a Redox Reaction. Specifically, as it produces a significant amount of heat, it is also known as a Thermite Reaction.

Step 3: Visual Representation

Final Solution: The given reaction is a Displacement Reaction (specifically a single displacement reaction), and more broadly, a Redox Reaction.

1. Theoretical Definition of Rancidity

Rancidity is defined as the process of slow oxidation of oil and fat present in food materials, resulting in a change in smell and taste. This phenomenon occurs when food items containing fats and oils are exposed to air (oxygen) for prolonged periods [Per the principles of Chemical Oxidation].

2. The Chemical Mechanism

The reaction involves the atmospheric oxygen ($O_2$) attacking the unsaturated fatty acid chains within food. This leads to the formation of volatile aldehydes, ketones, and carboxylic acids, which are responsible for the foul odor and rancid taste.

3. Illustrative Representation of Oxidative Degradation

4. Practical Example: Potato Chips

A quintessential example of rancidity is observed in deep-fried snacks such as potato chips. If left exposed to the atmosphere, the lipids in the chips react with oxygen, making them unfit for consumption. To prevent this, manufacturers employ specific preservation techniques:

5. Analytical Conclusion

Rancidity is essentially a form of chemical spoilage. By understanding that it is an oxidation reaction, we can scientifically control it by limiting exposure to oxidizing agents, lowering ambient temperature, or utilizing inert atmospheres [Based on the Rate of Reaction kinetics].

Final Solution: Rancidity is the oxidation of fats and oils in food, leading to unpleasant odors and tastes. Example: Potato chips becoming stale and foul-smelling when exposed to atmospheric oxygen over time.

Step 1: Analyzing the Unbalanced Equation

The given skeletal chemical equation is:

$NaCl_{(aq)} + AgNO_{3(aq)} \rightarrow AgCl_{(s)} + NaNO_{3(aq)}$

To balance this equation, we must ensure that the number of atoms for each element is equal on both the reactant and product sides [Per the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction].

Step 2: Tabulating Atomic Distribution

Step 3: Assessment and Verification

By inspecting the table above, we observe the following:

• The number of $Na$ atoms is 1 on both sides.
• The number of $Cl$ atoms is 1 on both sides.
• The number of $Ag$ atoms is 1 on both sides.
• The number of $NO_3$ groups is 1 on both sides.

Since the stoichiometry of every element and polyatomic ion is already balanced, the coefficients of all reactants and products are effectively $1$.

Step 4: Visual Representation of the Precipitation Reaction

Step 5: Final Equation Formulation

The equation remains as provided, confirming that the initial representation follows the stoichiometric requirement of balancing [Law of Conservation of Mass].

Final Solution: $1NaCl + 1AgNO_3 \rightarrow 1AgCl + 1NaNO_3$

Step 1: Identifying Chemical Formulas and States

To construct the chemical equation, we first translate the chemical names into their respective molecular or ionic formulas, incorporating their physical states as provided:

• Magnesium (solid): $Mg_{(s)}$
• Hydrochloric acid (aqueous): $HCl_{(aq)}$
• Magnesium chloride (aqueous): $MgCl_{2(aq)}$
• Hydrogen (gas): $H_{2(g)}$

Step 2: Constructing the Skeletal Equation

The unbalanced skeletal equation is written by placing reactants on the left and products on the right [Per the Law of Conservation of Mass]:

$Mg_{(s)} + HCl_{(aq)} \rightarrow MgCl_{2(aq)} + H_{2(g)}$

Step 3: Balancing the Chemical Equation

We apply the inspection method to ensure the number of atoms of each element is equal on both sides of the arrow.

Adjustment: To balance Chlorine and Hydrogen, place a coefficient of $2$ before $HCl$:

$Mg_{(s)} + 2HCl_{(aq)} \rightarrow MgCl_{2(aq)} + H_{2(g)}$

[Verification: Reactant side = 1 Mg, 2 H, 2 Cl; Product side = 1 Mg, 2 H, 2 Cl. The equation is now balanced.]

Step 4: Identification of Reaction Type

The reaction involves a more reactive metal (Magnesium) displacing a less reactive element (Hydrogen) from its acid compound. This is defined as a Single Displacement Reaction [Per the activity series of metals].

Final Solution: The balanced equation is $Mg_{(s)} + 2HCl_{(aq)} \rightarrow MgCl_{2(aq)} + H_{2(g)}$, and the reaction type is a Single Displacement Reaction.

Step 1: Identification of Reactants and Products

To translate the word equation into a chemical equation, we identify the chemical species involved:

• Reactants: Hydrogen gas ($H_2$) and Nitrogen gas ($N_2$). [Note: Both are diatomic molecules under standard conditions].
• Product: Ammonia ($NH_3$).

The unbalanced skeletal equation is:

$H_2(g) + N_2(g) \rightarrow NH_3(g)$

Step 2: Analysis of Atom Counts

Using the Law of Conservation of Mass [per Antoine Lavoisier], the number of atoms of each element must be equal on both sides of the arrow.

Step 3: Balancing the Chemical Equation

Part A: Balance Nitrogen atoms
To balance the Nitrogen atoms, place a coefficient of $2$ in front of $NH_3$ on the RHS.

$H_2 + N_2 \rightarrow 2NH_3$

Updated Count: RHS has $2 \times 3 = 6$ Hydrogen atoms and $2$ Nitrogen atoms.

Part B: Balance Hydrogen atoms
To achieve $6$ Hydrogen atoms on the LHS, place a coefficient of $3$ in front of $H_2$.

$3H_2 + N_2 \rightarrow 2NH_3$

Step 4: Verification of Atomic Balance

We perform a final audit of the balanced equation:

• LHS: $N = 2$, $H = (3 \times 2) = 6$
• RHS: $N = 2$, $H = (2 \times 3) = 6$

Since the number of atoms for each element is identical on both sides, the equation satisfies the Law of Conservation of Mass.

Final Solution: $3H_2(g) + N_2(g) \rightarrow 2NH_3(g)$

Step 1: Identification of Reactants and Products

The unbalanced chemical equation provided is: $NaOH + H_2SO_4 \rightarrow Na_2SO_4 + H_2O$

Following the Law of Conservation of Mass [Matter is neither created nor destroyed in a chemical reaction; therefore, the total number of atoms of each element must remain constant on both sides of the reaction].

Step 2: Balancing the Sodium ($Na$) Atoms

To balance the $Na$ atoms, place a coefficient of $2$ in front of $NaOH$ on the left-hand side [Trial and Error Method / Hit and Trial Method].

$2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + H_2O$

Step 3: Balancing Hydrogen ($H$) and Oxygen ($O$) Atoms

Recounting atoms after Step 2:

• $Na$: 2 on LHS, 2 on RHS (Balanced)
• $S$: 1 on LHS, 1 on RHS (Balanced)
• $H$: ($2 \times 1$) + 2 = 4 on LHS. To match 4 on RHS, place a coefficient of $2$ in front of $H_2O$.
• $O$: ($2 \times 1$) + 4 = 6 on LHS. On RHS, $4 + (2 \times 1) = 6$ (Balanced).

Step 4: Final Verification of Atoms

The balanced equation is: $2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$

Final Solution: 2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O

Step 1: Identification of Reactants and Products

Based on the chemical statement provided, we identify the substances involved:

• Hydrogen sulphide gas: $H_2S(g)$
• Air (Oxygen gas): $O_2(g)$
• Water: $H_2O(l)$
• Sulphur dioxide: $SO_2(g)$

[Per the Law of Conservation of Mass, the total mass of reactants must equal the total mass of products, necessitating a balanced chemical equation.]

Step 2: Writing the Unbalanced Skeleton Equation

Representing the reaction symbolically:

$H_2S + O_2 \rightarrow H_2O + SO_2$

Step 3: Atomic Accounting

Quantifying the number of atoms on both sides of the equation:

Step 4: Balancing the Atoms

To balance the Oxygen atoms, we first adjust the coefficients. Since we have 3 oxygen atoms on the RHS and 2 on the LHS, we apply a coefficient of $3$ to $O_2$ and $2$ to both products to balance the oxygen count:

$H_2S + 3O_2 \rightarrow 2H_2O + 2SO_2$

Now, adjust the Hydrogen and Sulphur counts to compensate for the coefficients applied to the products:

• LHS: $H=2, S=1$
• RHS: $H=4, S=2$

To balance the reactants with the new totals on the RHS, apply a coefficient of $2$ to $H_2S$:

$2H_2S + 3O_2 \rightarrow 2H_2O + 2SO_2$

Step 5: Final Verification

Re-counting the atoms in the balanced equation $2H_2S + 3O_2 \rightarrow 2H_2O + 2SO_2$:

• Hydrogen: $2 \times 2 = 4$ (LHS), $2 \times 2 = 4$ (RHS)
• Sulphur: $2 \times 1 = 2$ (LHS), $2 \times 1 = 2$ (RHS)
• Oxygen: $3 \times 2 = 6$ (LHS), $(2 \times 1) + (2 \times 2) = 6$ (RHS)

Final Solution: $2H_2S(g) + 3O_2(g) \rightarrow 2H_2O(l) + 2SO_2(g)$

Step 1: Chemical Analysis of the Reaction

The problem describes a chemical transformation wherein a shiny brown metallic element ($X$) is heated in the presence of atmospheric oxygen ($O_2$), resulting in a black-colored substance. This reaction is classified as a combination reaction and specifically an oxidation reaction of a metal [Per the Law of Conservation of Mass].

Step 2: Identification of Element 'X'

Copper ($Cu$) is a metal that is characteristically "shiny brown" in its elemental form. When subjected to heating in the presence of oxygen, it reacts to form copper(II) oxide. This observation matches the physical description provided in the prompt.

Step 3: Derivation of the Chemical Equation

The chemical reaction proceeds as follows:

$2Cu(s) + O_2(g) \xrightarrow{\Delta} 2CuO(s)$

Breakdown of the reaction:

• Reactants: Copper ($Cu$, shiny brown solid) and Oxygen ($O_2$, gaseous, from air).
• Product: Copper(II) Oxide ($CuO$, black solid).
• Process: The heat ($\Delta$) facilitates the addition of oxygen to the copper, resulting in the oxidation of copper to copper(II) oxide [By the Definition of Oxidation: Addition of Oxygen].

Step 4: Visual Representation of Oxidation

Step 5: Summary Table of Findings

Final Solution: The shiny brown element ‘$X$’ is Copper ($Cu$), and the black-colored compound formed is Copper(II) Oxide ($CuO$).

Step 1: Identification of Reactants and Products

To translate the word equation into a chemical equation, we first define the chemical formulas for each substance based on their ionic valencies:

• Barium chloride: $Ba^{2+}$ and $Cl^{-1}$ forms $BaCl_2$
• Aluminium sulphate: $Al^{3+}$ and $SO_4^{2-}$ forms $Al_2(SO_4)_3$
• Aluminium chloride: $Al^{3+}$ and $Cl^{-1}$ forms $AlCl_3$
• Barium sulphate: $Ba^{2+}$ and $SO_4^{2-}$ forms $BaSO_4$

Step 2: Skeletal Chemical Equation

Placing the reactants on the left and products on the right [Per the Law of Conservation of Mass]:

$BaCl_2(aq) + Al_2(SO_4)_3(aq) \rightarrow AlCl_3(aq) + BaSO_4(s)$

Step 3: Balancing the Chemical Equation

We apply the trial and error method to equate the number of atoms of each element on both sides:

Balancing Process:

1. Balance $SO_4$ groups: Place a coefficient of $3$ before $BaSO_4$ on the RHS. This results in $3$ Ba atoms on the RHS.
2. Balance $Ba$ atoms: Place a coefficient of $3$ before $BaCl_2$ on the LHS.
3. Balance $Al$ atoms: Place a coefficient of $2$ before $AlCl_3$ on the RHS.
4. Verification: LHS has $(3 \times 2) = 6$ Cl atoms and RHS has $(2 \times 3) = 6$ Cl atoms. The equation is now balanced.

Step 4: Balanced Chemical Equation

$3BaCl_2(aq) + Al_2(SO_4)_3(aq) \rightarrow 2AlCl_3(aq) + 3BaSO_4(s)$

Final Solution: 3BaCl_2(aq) + Al_2(SO_4)_3(aq) \rightarrow 2AlCl_3(aq) + 3BaSO_4(s)

Step 1: Identifying the Skeletal Equation

The given chemical equation represents a double displacement reaction between Barium chloride and Sulfuric acid:

$BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + HCl$

[Law of Conservation of Mass: Matter is neither created nor destroyed in a chemical reaction; therefore, the total mass of atoms on the reactant side must equal the total mass of atoms on the product side.]

Step 2: Tabulating Atomic Counts

We analyze the number of atoms of each element on both sides of the equation to identify the imbalance.

Step 3: Balancing the Imbalanced Elements

From the table above, the elements $Cl$ and $H$ are unbalanced. Both have 2 atoms on the LHS and 1 atom on the RHS.

To balance the chlorine ($Cl$) and hydrogen ($H$) atoms, we place a coefficient of $2$ in front of the hydrochloric acid ($HCl$) molecule on the product side. [By Trial and Error Method/Hit-and-Trial Method].

$BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + 2HCl$

Step 4: Final Verification

Re-calculating the atom count for the balanced equation:

• $Ba$: $1$ (LHS) = $1$ (RHS)
• $Cl$: $2$ (LHS) = $2 \times 1 = 2$ (RHS)
• $H$: $2$ (LHS) = $2 \times 1 = 2$ (RHS)
• $S$: $1$ (LHS) = $1$ (RHS)
• $O$: $4$ (LHS) = $4$ (RHS)

Visual Representation of the Balanced Reaction

Final Solution: $BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + 2HCl$

Theoretical Foundation: Definition of Reduction

In the context of classical chemical theory, reduction is defined as the loss of oxygen from a substance during a chemical reaction. Conversely, it is often paired with oxidation (gain of oxygen), forming a combined redox process. The general chemical schema is represented as follows:

$XY + Z \rightarrow X + YZ$ (where the species $XY$ loses oxygen to become $X$).

Example 1: Reduction of Copper(II) Oxide

When Copper(II) oxide is heated in a stream of hydrogen gas, the oxygen is removed from the copper, resulting in the formation of metallic copper and water vapor.

Chemical Equation:

$CuO(s) + H_2(g) \xrightarrow{\Delta} Cu(s) + H_2O(g)$

• Analysis: $CuO$ (Copper(II) oxide) loses its oxygen atom to become $Cu$ (Copper). [Per the classical definition of reduction].
• Observation: The black powder ($CuO$) turns into a reddish-brown metal ($Cu$).

Example 2: Reduction of Zinc Oxide

When Zinc oxide is heated with carbon (coke), the carbon acts as a reducing agent, effectively stripping the oxygen from the zinc oxide.

Chemical Equation:

$ZnO(s) + C(s) \xrightarrow{\Delta} Zn(s) + CO(g)$

• Analysis: $ZnO$ (Zinc oxide) loses oxygen to become $Zn$ (Zinc). [Per the classical definition of reduction].
• Observation: The white solid ($ZnO$) is reduced to metallic zinc vapor/solid.

Visual Representation of Reductive Oxygen Loss

Comparative Analysis Table

Final Solution: Reduction is defined as the loss of oxygen from a chemical species. As demonstrated, $CuO$ is reduced to $Cu$ by losing oxygen to hydrogen, and $ZnO$ is reduced to $Zn$ by losing oxygen to carbon.

1. Theoretical Definition of Corrosion

Corrosion is defined as the natural, electrochemical process by which a refined metal is gradually converted into a more chemically stable form, such as an oxide, hydroxide, or sulfide. This phenomenon occurs when a metal surface is exposed to environmental agents like oxygen, moisture (water vapor), acids, or salts. [Per the principles of Electrochemistry and Redox Reactions].

2. Mechanism and Chemical Basis

Corrosion is a spontaneous oxidation reaction. The metal atoms lose electrons to form metal ions at the anode site, while oxygen is reduced at the cathode site. This leads to the formation of a metal compound that typically lacks the mechanical strength and luster of the original metallic element.

3. Exemplification: The Rusting of Iron

The most ubiquitous example of corrosion is the rusting of iron ($Fe$). When iron objects are exposed to moist air, they react with oxygen in the presence of water to form hydrated iron(III) oxide, commonly known as rust.

Chemical Reaction:

$4Fe_{(s)} + 3O_{2(g)} + 2xH_2O_{(l)} \rightarrow 2Fe_2O_3 \cdot xH_2O_{(s)}$

4. Comparative Analysis of Factors

5. Prevention Strategies

To inhibit the corrosive process, several industrial and practical methods are employed:

• Galvanization: Coating iron with a thin layer of zinc. Zinc is more reactive than iron and preferentially oxidizes, protecting the underlying metal [Per the Electrochemical Series].
• Painting/Coating: Creating a physical barrier between the metal surface and the atmosphere.
• Alloying: Mixing metals (e.g., Stainless Steel = $Fe + Cr + Ni$) to enhance chemical resistance.

Final Solution: Corrosion is the gradual degradation of a metal due to chemical reactions with its environment, characterized by the oxidation of the metal; the rusting of iron, represented by the equation $4Fe + 3O_2 + 2xH_2O \rightarrow 2Fe_2O_3 \cdot xH_2O$, serves as the primary example.

1. Theoretical Foundation: The Mechanism of Corrosion

The application of paint on iron articles serves as a protective measure against corrosion, specifically the process known as rusting. Iron ($Fe$) reacts spontaneously with atmospheric oxygen ($O_2$) and moisture ($H_2O$) to form hydrated iron(III) oxide ($Fe_2O_3 \cdot xH_2O$).

The chemical reaction is represented as follows:

$4Fe(s) + 3O_2(g) + 2xH_2O(l) \rightarrow 2Fe_2O_3 \cdot xH_2O(s)$

[Per the Principles of Redox Chemistry, iron acts as the reducing agent and oxygen as the oxidizing agent in this electrochemical process.]

2. The Protective Mechanism of Painting

Painting functions as a physical barrier method. By coating the surface of the iron, the paint creates an impermeable layer that prevents the contact of the metal substrate with the surrounding environment.

• Exclusion of Reactants: The paint layer blocks the diffusion of oxygen molecules to the iron surface.
• Moisture Sealing: It prevents water vapor or liquid water from reaching the iron, thereby inhibiting the electrolytic medium required for the corrosion cell to function.
• Prevention of Rust Propagation: Without the electrochemical circuit (anode: $Fe \rightarrow Fe^{2+} + 2e^-$; cathode: $O_2 + 2H_2O + 4e^- \rightarrow 4OH^-$), the oxidation of iron is halted at the interface.

3. Visual Representation of Corrosion Inhibition

4. Comparative Summary of Protective Measures

Final Solution: We apply paint on iron articles to create a physical barrier that prevents the metal from coming into contact with atmospheric oxygen and moisture, thereby arresting the chemical process of corrosion (rusting) and extending the structural integrity and lifespan of the iron.

Step 1: Identification of Reactants and Products

Based on the provided chemical description, the reaction involves the displacement of copper from a solution of copper(II) chloride by aluminium. The chemical entities are identified as follows:

• Aluminium: $Al$ (Solid)
• Copper(II) Chloride: $CuCl_2$ (Aqueous)
• Aluminium Chloride: $AlCl_3$ (Aqueous)
• Copper: $Cu$ (Solid)

Step 2: Skeletal Chemical Equation

The unbalanced skeletal equation representing this displacement reaction is:

$Al_{(s)} + CuCl_{2(aq)} \rightarrow AlCl_{3(aq)} + Cu_{(s)}$

Step 3: Balancing the Atoms

To satisfy the [Law of Conservation of Mass], the number of atoms of each element must be equal on both sides of the equation. We analyze the stoichiometry:

To balance the chlorine atoms, we find the least common multiple of 2 and 3, which is 6. Multiply $CuCl_2$ by 3 and $AlCl_3$ by 2:

$Al + 3CuCl_2 \rightarrow 2AlCl_3 + Cu$

Now, adjust the remaining atoms. We have 2 atoms of $Al$ on the RHS, so we place a coefficient of 2 before $Al$ on the LHS. We have 3 atoms of $Cu$ on the LHS, so we place a coefficient of 3 before $Cu$ on the RHS.

Step 4: Final Balanced Equation

The final balanced chemical equation is:

$2Al_{(s)} + 3CuCl_{2(aq)} \rightarrow 2AlCl_{3(aq)} + 3Cu_{(s)}$

Visual Representation of the Displacement Process

Theoretical Justification: This is classified as a single displacement reaction [Per the Activity Series of Metals], where aluminium (a more reactive metal) displaces copper (a less reactive metal) from its salt solution.

Final Solution: 2Al_{(s)} + 3CuCl_{2(aq)} \rightarrow 2AlCl_{3(aq)} + 3Cu_{(s)}