Question

Balance each of the following chemical equations. a. \(\mathrm{C}_{2}(g)+\mathrm{KBr}(a q) \rightarrow \mathrm{Br}_{2}(I)+\mathrm{KCl}(a q)\) b. \(\operatorname{Cr}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(s)\) c. \(P_{4}(s)+H_{2}(g) \rightarrow P H_{3}(g)\) d. \(\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)\) e. \(\mathrm{PCl}_{3}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+\mathrm{HCl}(a q)\) f. \(\mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{3}(g)\) g. \(\mathrm{C}_{7} \mathrm{H}_{16}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) h. \(C_{2} H_{6}(g)+O_{2}(g) \rightarrow C O_{2}(g)+H_{2} O(g)\)

Short answer

a. C₂(g) + 2 KBr(aq) → Br₂(l) + 2 KCl(aq) b. 2 Cr(s) + O₂(g) → Cr₂O₃(s) c. P₄(s) + 6 H₂(g) → 4 PH₃(g) d. 2 Al(s) + 3 H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3 H₂(g) e. PCl₃(l) + 3 H₂O(l) → H₃PO₃(aq) + 3 HCl(aq) f. 2 SO₂(g) + O₂(g) → 2 SO₃(g) g. C₇H₁₆(l) + 11 O₂(g) → 7 CO₂(g) + 8 H₂O(g) h. 2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(g)

Step-by-step solution

a. Balancing C₂(g) + KBr(aq) → Br₂(l) + KCl(aq)

We start by listing the number of atoms of each element on both sides of the equation: - Carbon: 2 on the left, 0 on the right - Potassium: 1 on the left, 1 on the right - Bromine: 1 on the left, 2 on the right - Chlorine: 0 on the left, 1 on the right To balance the equation: 1. Add a coefficient of 2 in front of KCl: C₂(g) + KBr(aq) → Br₂(l) + 2 KCl(aq) 2. Add a coefficient of 2 in front of KBr: C₂(g) + 2 KBr(aq) → Br₂(l) + 2 KCl(aq) Now the equation is balanced.

b. Balancing Cr(s) + O₂(g) → Cr₂O₃(s)

To balance the equation: 1. Add a coefficient of 2 in front of Cr: 2 Cr(s) + O₂(g) → Cr₂O₃(s) Now the equation is balanced.

c. Balancing P₄(s) + H₂(g) → PH₃(g)

To balance the equation: 1. Add a coefficient of 4 in front of PH₃: P₄(s) + H₂(g) → 4 PH₃(g) 2. Add a coefficient of 6 in front of H₂: P₄(s) + 6 H₂(g) → 4 PH₃(g) Now the equation is balanced.

d. Balancing Al(s) + H₂SO₄(aq) → Al₂(SO₄)₃(aq) + H₂(g)

To balance the equation: 1. Add a coefficient of 2 in front of Al: 2 Al(s) + H₂SO₄(aq) → Al₂(SO₄)₃(aq) + H₂(g) 2. Add a coefficient of 3 in front of H₂SO₄: 2 Al(s) + 3 H₂SO₄(aq) → Al₂(SO₄)₃(aq) + H₂(g) 3. Add a coefficient of 3 in front of H₂: 2 Al(s) + 3 H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3 H₂(g) Now the equation is balanced.

e. Balancing PCl₃(l) + H₂O(l) → H₃PO₃(aq) + HCl(aq)

To balance the equation: 1. Add a coefficient of 3 in front of HCl: PCl₃(l) + H₂O(l) → H₃PO₃(aq) + 3 HCl(aq) 2. Add a coefficient of 3 in front of H₂O: PCl₃(l) + 3 H₂O(l) → H₃PO₃(aq) + 3 HCl(aq) Now the equation is balanced.

f. Balancing SO₂(g) + O₂(g) → SO₃(g)

To balance the equation: 1. Add a coefficient of 2 in front of SO₃: SO₂(g) + O₂(g) → 2 SO₃(g) 2. Add a coefficient of 2 in front of SO₂: 2 SO₂(g) + O₂(g) → 2 SO₃(g) Now the equation is balanced.

g. Balancing C₇H₁₆(l) + O₂(g) → CO₂(g) + H₂O(g)

To balance the equation: 1. Add a coefficient of 7 in front of CO₂: C₇H₁₆(l) + O₂(g) → 7 CO₂(g) + H₂O(g) 2. Add a coefficient of 8 in front of H₂O: C₇H₁₆(l) + O₂(g) → 7 CO₂(g) + 8 H₂O(g) 3. Add a coefficient of 11 in front of O₂: C₇H₁₆(l) + 11 O₂(g) → 7 CO₂(g) + 8 H₂O(g) Now the equation is balanced.

h. Balancing C₂H₆(g) + O₂(g) → CO₂(g) + H₂O(g)

To balance the equation: 1. Add a coefficient of 2 in front of CO₂: C₂H₆(g) + O₂(g) → 2 CO₂(g) + H₂O(g) 2. Add a coefficient of 3 in front of H₂O: C₂H₆(g) + O₂(g) → 2 CO₂(g) + 3 H₂O(g) 3. Add a coefficient of 7/2 in front of O₂: C₂H₆(g) + 7/2 O₂(g) → 2 CO₂(g) + 3 H₂O(g) Alternatively, to avoid fractional coefficients, we can multiply the entire equation by 2: C₂H₆(g) + 3.5 O₂(g) → 2 CO₂(g) + 3 H₂O(g) becomes 2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(g) Now the equation is balanced.

Key concepts

Stoichiometry

Stoichiometry is an essential concept in chemistry, focusing on the relationship between the quantities of reactants and products in a chemical reaction. When balancing chemical equations, stoichiometry ensures that the number of atoms for each element is the same on both sides of the equation. This is crucial because it maintains the integrity of the chemical process and enables us to predict the quantities needed and those that will be produced.

In order to apply stoichiometry effectively, you must first comprehend the coefficients in a balanced chemical equation. These coefficients indicate the proportion of moles, molecules, or volumes in which the substances react. Understanding these proportions lets us use stoichiometry to solve real-world problems, such as calculating how much of a reactant is needed to produce a certain amount of product.

Quick facts about stoichiometry:

• It utilizes ratios derived from chemical equations.
• It allows conversion between moles, grams, and particles.
• Helps determine limiting reactants and theoretical yield.
• Fundamental for quantitative chemical analysis.

Chemical Reactions

Chemical reactions involve the breaking of bonds in reactants and the formation of new bonds to create products. A balanced chemical equation represents this transformation. Different types of chemical reactions, such as synthesis, decomposition, single replacement, and combustion, have unique characteristics and patterns.

In a balanced equation, both the type and number of atoms remain constant between reactants and products, which is a fundamental characteristic of all chemical reactions. Understanding the different types of reactions helps predict product formation and energy changes. For example:

• Combustion: A reaction where a substance combines with oxygen, releasing heat and often forming carbon dioxide and water, such as the burning of hydrocarbons like ethane or hexane.
• Synthesis: Two or more simple substances combine to form a more complex product, such as forming aluminum sulfate from aluminum and sulfuric acid.
• Decomposition: A complex molecule breaks down into simpler products, often involving heating or other energy input.

Recognizing how reactants transform into products allows you to construct and balance chemical reactions effectively and accurately.

Law of Conservation of Mass

The Law of Conservation of Mass states that mass is conserved in a chemical reaction; it cannot be created or destroyed. This law is the foundation for balancing chemical equations. It implies that the total mass of reactants equals the total mass of products. Therefore, all atoms present in reactants must be accounted for in the products.

In practical terms, when balancing a chemical equation, each type of atom must appear in the same quantity on both sides of the equation. If any discrepancies exist, the equation is unbalanced, indicating a violation of this fundamental law. For example, as seen in the balanced equation of a combustion reaction like C₂H₆ and O₂ forming CO₂ and H₂O, each carbon and oxygen atom present in the reactants corresponds exactly to those in the products.

Balancing a chemical equation also honors this principle, ensuring that all matter is accounted for through proper stoichiometric coefficients. This process validates our calculations and predictions about how substances will behave and combine during reactions.

Balancing Methods

Balancing chemical equations is a skill that involves several methods to ensure all elements have the same number of atoms on both sides of the reaction. Here are some common techniques for balancing equations efficiently:

• Inspection Method: This involves adding coefficients to adjust the number of atoms of each element, starting with one that appears in a single reactant and product. Adjust larger, more complex molecules last.
• Algebraic Method: Helps in balancing complex reactions by setting up algebraic equations based on the number of each type of atom.
• Oxidation Number Method: Often used for redox reactions, it focuses on the change in oxidation states to achieve balance.

Regardless of the method, the goal is to balance the equation so that it satisfies the Law of Conservation of Mass. During this process, keep in mind:

• You can only adjust coefficients (numbers in front of molecules), not subscripts within formulas.
• Double-check atom numbers for populous elements in reactants and products.
• Balancing coefficients should be the smallest set of integers possible.

Understanding these processes through practice will make balancing chemical equations straightforward and intuitive.