Problem 11 - Chemistry: The Central Science

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) \(\mathrm{ZnO}(\mathrm{s})+\mathrm{CO}(g) \stackrel{\Delta}{\longrightarrow}\)

Step-by-Step Solution

Answer

(a) 2Cr₂O₃(s) + 8Na(l) ⟶ 4Na₂CrO₄(s) (b) PbCO₃(s) ⟶ PbO(s) + CO₂(g) (c) 2CdS(s) + 3O₂(g) ⟶ 2CdO(s) + 2SO₂(g) (d) ZnO(s) + CO(g) ⟶ Zn(s) + CO₂(g)

Step 1: Identify the Products

Here, Chromium(III) oxide (Cr₂O₃) reacts with Sodium (Na) to produce sodium chromate (Na₂CrO₄) as the reaction proceeds. So the unbalanced equation is: Cr₂O₃(s) + Na(l) ⟶ Na₂CrO₄ (s)

Step 2: Balance the Equation

By comparing the reactants and products, we can balance the equation: 2Cr₂O₃(s) + 8Na(l) ⟶ 4Na₂CrO₄(s) (b) PbCO₃(s) ⟶

Step 3: Identify the Products

When lead(II) carbonate (PbCO₃) is heated (Δ), it decomposes into lead(II) oxide (PbO) and carbon dioxide (CO₂). So the unbalanced equation is: PbCO₃(s) ⟶ PbO(s) + CO₂(g)

Step 4: Balance the Equation

The equation is already balanced: PbCO₃(s) ⟶ PbO(s) + CO₂(g) (c) CdS(s) + O₂(g) ⟶

Step 5: Identify the Products

When cadmium sulfide (CdS) reacts with oxygen (O₂) upon heating (Δ), it forms cadmium oxide (CdO) and sulfur dioxide (SO₂). So the unbalanced equation is: CdS(s) + O₂(g) ⟶ CdO(s) + SO₂(g)

Step 6: Balance the Equation

By comparing the reactants and products, we can balance the equation: 2CdS(s) + 3O₂(g) ⟶ 2CdO(s) + 2SO₂(g) (d) ZnO(s) + CO(g) ⟶

Step 7: Identify the Products

When zinc oxide (ZnO) reacts with carbon monoxide (CO) upon heating (Δ), it forms zinc and carbon dioxide (CO₂). So the unbalanced equation is: ZnO(s) + CO(g) ⟶ Zn(s) + CO₂(g)

Step 8: Balance the Equation

The equation is already balanced: ZnO(s) + CO(g) ⟶ Zn(s) + CO₂(g)

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a crucial step when working with chemical reactions, ensuring the law of conservation of mass is upheld. This means the number of each type of atom must be the same on both sides of the equation. For example, in the reaction between chromium(III) oxide and sodium:

• The unbalanced equation: \(\mathrm{Cr}_{2} \mathrm{O}_{3}(s) + \mathrm{Na}(l) \longrightarrow \mathrm{Na}_{2}\mathrm{CrO}_{4}(s)\)

By inspecting and adding coefficients, we achieve:

• Balanced equation: \(2\mathrm{Cr}_{2}\mathrm{O}_{3}(s) + 8\mathrm{Na}(l) \longrightarrow 4\mathrm{Na}_{2}\mathrm{CrO}_{4}(s)\)

The coefficients ensure there are equal numbers of Cr, Na, and O atoms on both sides.
Practice makes perfect when balancing equations, starting with individual elements and adjusting coefficients until the equation is balanced.

Decomposition Reactions

Decomposition reactions occur when a compound breaks down into simpler substances, often assisted by heating. This is often represented by the symbol \(\Delta\) over the arrow. A good example is lead(II) carbonate, \(\mathrm{PbCO}_{3}\), breaking down to form lead(II) oxide \(\mathrm{PbO}\) and carbon dioxide \(\mathrm{CO}_{2}\):

• Unbalanced: \(\mathrm{PbCO}_{3}(s) \stackrel{\Delta}{\longrightarrow} \mathrm{PbO}(s) + \mathrm{CO}_{2}(g)\)
• Balanced: Actually, it is already balanced: \(\mathrm{PbCO}_{3}(s) \stackrel{\Delta}{\longrightarrow} \mathrm{PbO}(s) + \mathrm{CO}_{2}(g)\)

Decomposition reactions are foundational in chemistry and help understand how substances interact with energy, often needing specific conditions like heat to proceed.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between chemical species. These reactions are crucial because they describe fundamental processes in chemistry, such as those related to energy production and storage. Let's examine the reaction of cadmium sulfide \(\mathrm{CdS}(s)\) with oxygen:

• Unbalanced: \(\mathrm{CdS}(s) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{CdO}(s) + \mathrm{SO}_{2}(g)\)
• Balanced: \(2\mathrm{CdS}(s) + 3\mathrm{O}_{2}(g) \longrightarrow 2\mathrm{CdO}(s) + 2\mathrm{SO}_{2}(g)\)

In this reaction:
- Cadmium is oxidized as it forms cadmium oxide.
- Sulfur is reduced to form sulfur dioxide. Understanding which elements undergo oxidation and reduction helps explain the mechanics behind resourceful chemical interactions.

Stoichiometry

Stoichiometry provides the quantitative relationship between reactants and products in a chemical reaction. It's a method used to calculate the masses, moles, and possibly, volumes of reactants and products. Consider the reaction between zinc oxide \(\mathrm{ZnO}\) and carbon monoxide \(\mathrm{CO}\):

• Unbalanced: \(\mathrm{ZnO}(s) + \mathrm{CO}(g) \longrightarrow \mathrm{Zn}(s) + \mathrm{CO}_{2}(g)\)
• Balanced: remains balanced as \(\mathrm{ZnO}(s) + \mathrm{CO}(g) \longrightarrow \mathrm{Zn}(s) + \mathrm{CO}_{2}(g)\)

Stoichiometry aids in determining:
- The amount of zinc produced from a given amount of zinc oxide.
- The mass of carbon monoxide required. Understanding these relationships is vital for practical applications in lab settings and industrial processes, ensuring sufficient reactants and predicting product yield.