General Chemistry Exam 2 Study Guide CH.4-6
GENERAL CHEMISTRY Exam 2 Study Guide: Chapters 4-6
Chapter 4: Chemical Reactions and Quantities
4.2 Writing and Balancing Chemical Equations
Chemical Reaction Notation: A chemical reaction can be interpreted in terms of moles or molecules.
Example: ext{O}2 (g) + 2 ext{H}2 (g)
ightarrow 2 ext{H}_2 ext{O} (l)1 mole of ext{O}_2
2 moles of ext{H}_2
2 moles of ext{H}_2 ext{O}
Balancing Chemical Equations:
Write and balance equations by manipulating coefficients to ensure the number of each type of atom is the same on both sides.
Always balance atoms in compounds before balancing free elements.
Balance atoms that occur as free elements last.
If fractions arise in balancing, multiply through by the least common denominator.
Example of Balancing:
Initial Reaction: ext{H}2 ext{SO}4 (aq) + ext{Ba(NO}3)2 (aq)
ightarrow ext{HNO}3 (aq) + ext{BaSO}4 (s)Balanced Reaction: ext{H}2 ext{SO}4 (aq) + ext{Ba(NO}3)2 (aq)
ightarrow 2 ext{HNO}3 (aq) + ext{BaSO}4 (s)
Example with Sodium Bicarbonate:
Reaction: ext{Na}2 ext{CO}3 (s) + 2 ext{HBr} (aq)
ightarrow 2 ext{NaBr} (aq) + ext{H}2 ext{O} (l) + ext{CO}2 (g)
4.3 Stoichiometry
Stoichiometry: Use stoichiometric ratios from a balanced equation to determine amounts of reactants consumed and products formed.
Example: ext{CH}4 + 2 ext{O}2
ightarrow ext{CO}2 + 2 ext{H}2 ext{O}Stoichiometric Ratio: 1:1:2 (for ext{CH}4 to ext{CO}2 and ext{H}_2 ext{O})
Moles and Mass Relationship:
Moles = rac{ ext{mass (g)}}{ ext{molar mass (g/mol)}}
Mass = Moles * Molar Mass
4.4 Stoichiometric Relations
Key Concepts:
Limiting Reactant: The reactant that will be completely consumed first, limiting the amount of products formed.
Theoretical Yield: The maximum amount of product that can be formed from given amounts of reactants.
% Yield: A measure of the efficiency of a reaction. Defined as $ ext{% yield} = rac{ ext{actual yield}}{ ext{theoretical yield}} imes 100$.
Example of Finding Limiting Reactant:
Reaction: ext{N}2 + 3 ext{H}2
ightarrow 2 ext{N}2 ext{H}4Given: 5 moles of ext{H}2 and 3 moles of ext{O}2
Determine Limiting Reactant by comparing amounts required based on stoichiometric ratio.
Calculate Theoretical Yield using proper stoichiometric conversion.
4.5 Combustion, Alkali Metals, and Halogens
Combustion Reactions: Typically involve a hydrocarbon reacting with oxygen to produce carbon dioxide and water.
General Equation: ext{Hydrocarbon} + ext{O}2 ightarrow ext{CO}2 + ext{H}_2 ext{O}
Process: Balance carbon atoms first, then hydrogen, followed by oxygen.
Example: For propane combustion, ext{C}3 ext{H}8 + 5 ext{O}2 ightarrow 3 ext{CO}2 + 4 ext{H}_2 ext{O}
Chapter 5: Solutions and Aqueous Reactions
5.2 Solution Concentration
Definition: A solution is a homogeneous mixture of two or more substances with uniform composition throughout.
Solvent: The substance in a solution present in the largest amount, which dissolves other substances.
Solute: The substance dissolved in the solvent.
Aqueous Solution: When water is the solvent.
Molarity (M): A measure of concentration defined as moles of solute per liter of solution: M = rac{n}{V}
Dilution Formula: For dilutions, the formula is: M1 V1 = M2 V2.
Example of Dilution Calculation: 200 mL of 8.25 M HCl diluted to find volume of a more dilute solution.
Solve for V2: V2 = rac{M1V1}{M_2} = rac{8.25 imes 200}{12.6}
ightarrow ext{Calculate}.
5.3 Solution Stoichiometry
Stoichiometry in Aqueous Solutions: Use molarity to determine relative amounts of reactants and products based on volume and concentrations.
Example calculation: Given 32 g of zinc reacts with hydrochloric acid, determine the volume of HCl needed if concentration is known.
Reaction: ext{Zn} + 2 ext{HCl}
ightarrow ext{ZnCl}2 + ext{H}2
Example of Complete Calculation: 6.0 M HCl + a specified quantity of zinc.
Convert grams of Zn to moles and then use stoichiometric ratios to find volume of HCl (in mL).
5.4 Types of Aqueous Solutions and Solubility
Water as a Solvent:
Water is a polar molecule; it effectively dissolves ionic compounds due to its polar nature, allowing ions to separate and interact with the solvent.
Electrolytes:
Strong Electrolytes: Compounds that completely dissociate in solution (Examples: NaCl, HCl, KOH).
Weak Electrolytes: Compounds that only partially dissociate (Examples: Vinegar, acetic acid, NH3).
Non-Electrolytes: Do not ionize or conduct electricity (Example: Sugar).
Acids and Bases:
Strong Acids: Completely dissociate in water (Examples: HCl, HNO3).
Weak Acids: Partially dissociate in water (Examples: acetic acid).
Strong Bases: Completely ionize in water (Example: NaOH).
Weak Bases: Partially dissociate in water (Example: NH3).
5.5 Precipitation Reactions
Definition: A precipitation reaction occurs when two aqueous solutions are mixed, leading to the formation of an insoluble solid, known as a precipitate.
Predicting Products: Use solubility rules to determine if a precipitate forms.
General Reaction Format: ext{AB}(aq) + ext{CD}(aq)
ightarrow ext{AD}(aq) + ext{CB}(s).
Example of Precipitation Reaction:
ext{KCl}(aq) + ext{AgNO}3(aq) ightarrow ext{KNO}3(aq) + ext{AgCl}(s).
5.6 Molecular, Ionic, and Net Ionic Equations
Total Ionic Equation: A representation that shows all of the reactants and products in their aqueous forms, fully separated into ions.
Example: ext{AgNO}3(aq) + ext{NaCl}(aq) ightarrow ext{AgCl}(s) + ext{NaNO}3(aq) converts to:
ext{Ag}^+(aq) + ext{NO}3^-(aq) + ext{Na}^+(aq) + ext{Cl}^-(aq) ightarrow ext{AgCl}(s) + ext{Na}^+(aq) + ext{NO}3^-(aq).
Net Ionic Equation: This equation only includes the ions and molecules that form the precipitate or undergo a change.
Remove spectator ions (ions that appear in the same form on both sides of the equation).
Example: From the previous, remove ext{Na}^+ and ext{NO}_3^-, giving: ext{Ag}^+(aq) + ext{Cl}^-(aq)
ightarrow ext{AgCl}(s).