Chemistry 1110 Exam 3 Study Guide: Classification & Balancing of Chemical Reactions and Mole & Mass Relationships

Chapter 5: Classification and Balancing of Chemical Reactions

Writing and Balancing Chemical Equations

Chemical equations represent chemical reactions, showing reactants and products. Balancing ensures the conservation of mass and charge.

• Correct Chemical Formulas: Write formulas for each compound using subscripts to indicate the number of atoms.

• Balancing Equations: Use coefficients to balance the number of atoms of each element on both sides of the equation.

• Physical States: Indicate the physical state of each substance: (s) for solid, (l) for liquid, (g) for gas, (aq) for aqueous.

• Example: $\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(l)}$

Net Ionic Equations

Net ionic equations show only the species that participate in the reaction, omitting spectator ions.

• Spectator Ions: Ions that do not change during the reaction.

• Example: $\mathrm{Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)}$

Types of Chemical Reactions

Three major types of reactions commonly discussed:

• Combination (Synthesis): Two or more substances combine to form one product. Example: $\mathrm{2Na(s) + Cl_2(g) \rightarrow 2NaCl(s)}$

• Decomposition: One substance breaks down into two or more products. Example: $\mathrm{2H_2O(l) \rightarrow 2H_2(g) + O_2(g)}$

• Single Replacement: One element replaces another in a compound. Example: $\mathrm{Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s)}$

Solubility Rules

Solubility rules help predict whether a compound will dissolve in water or form a precipitate.

• Soluble Compounds: Most salts containing Na+, K+, NH4+, NO3-, and ClO4- are soluble.

• Insoluble Compounds: Most salts containing Ag+, Pb2+, and Hg22+ are insoluble.

• Application: Use solubility rules to predict precipitation reactions.

• Additional info: Refer to textbook pages 141-142 for detailed solubility rules.

Redox Reactions

Redox (reduction-oxidation) reactions involve the transfer of electrons between species.

• Oxidation: Loss of electrons; increase in oxidation state.

• Reduction: Gain of electrons; decrease in oxidation state.

• Oxidizing Agent: Causes oxidation; is itself reduced.

• Reducing Agent: Causes reduction; is itself oxidized.

• Example: $\mathrm{2Fe^{2+} + Cl_2 \rightarrow 2Fe^{3+} + 2Cl^-}$

• Key Point: Oxidation and reduction always occur together.

Key Terms

Review definitions of key terms related to reactions and balancing (see page 157 for textbook definitions).

• Reactant: Substance consumed in a reaction.

• Product: Substance formed in a reaction.

• Coefficient: Number in front of a formula indicating quantity.

• Subscript: Number within a formula indicating number of atoms.

• Spectator Ion: Ion not involved in the reaction.

• Precipitate: Insoluble product formed in a reaction.

Chapter 6: Chemical Reactions: Mole and Mass Relationships

Moles and Avogadro’s Number

The mole is a fundamental unit for counting particles in chemistry. Avogadro’s number relates moles to individual particles.

• Mole: One mole contains $6.022 \times 10^{23}$ particles (atoms, molecules, ions).

• Avogadro’s Number: $6.022 \times 10^{23}$

• Example: 2 moles of H2O contains $2 \times 6.022 \times 10^{23}$ molecules.

Molar Mass

Molar mass is the mass of one mole of a substance, used for conversions between mass and moles.

• Calculation: Sum the atomic masses of all atoms in a formula.

• Unit: grams per mole (g/mol)

• Example: Molar mass of H2O = $2 \times 1.01 + 16.00 = 18.02$ g/mol

Conversions Between Grams, Moles, and Molecules

Use molar mass and Avogadro’s number to convert between mass, moles, and number of particles.

• Grams to Moles: $\text{moles} = \frac{\text{grams}}{\text{molar mass}}$

• Moles to Molecules: $\text{molecules} = \text{moles} \times 6.022 \times 10^{23}$

• Example: 36.04 g H2O = $\frac{36.04}{18.02} = 2$ moles

Stoichiometry

Stoichiometry involves quantitative relationships between reactants and products in a chemical reaction.

• Balanced Equation: Use coefficients to relate moles of reactants and products.

• Conversion Steps: grams → moles → moles (other substance) → grams

• Example: $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$: 4 moles H2 produce 4 moles H2O.

Limiting Reagents

The limiting reagent is the reactant that determines the maximum amount of product formed.

• Identification: Calculate the amount of product each reactant can produce; the smallest is the limiting reagent.

• Example: If 5 g A and 10 g B are mixed, calculate product from each; whichever produces less is limiting.

Percent Yield

Percent yield compares the actual amount of product obtained to the theoretical maximum.

• Actual Yield: Amount of product actually obtained from experiment.

• Theoretical Yield: Maximum possible amount calculated from stoichiometry.

• Percent Yield Formula:

$\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

• Example: If theoretical yield is 10 g and actual yield is 8 g, percent yield is $\frac{8}{10} \times 100\% = 80\%$

Key Terms

Review definitions of key terms related to mole and mass relationships (see page 179 for textbook definitions).

• Mole: Unit for amount of substance.

• Molar Mass: Mass of one mole of a substance.

• Stoichiometry: Calculation of reactant and product quantities.

• Limiting Reagent: Reactant that limits product formation.

• Percent Yield: Ratio of actual to theoretical yield.

Additional info: Practice problems from the textbook are highly recommended for mastering these concepts.