Chemical Quantities and Aqueous Reactions

Chapter Four: Chemical Quantities and Aqueous Reactions

Overview of Chemical Reactions

• The amount of substances used and produced in a chemical reaction adheres to a quantitative relationship.

• Law of Conservation of Mass: Mass is conserved in a chemical reaction; the total mass of reactants equals the total mass of products.

• Balancing Equations: Requires adjusting the number of atoms of each element to be equal on both sides of the equation.

• The study of numerical relationships between chemical quantities in reactions is referred to as stoichiometry.

Reaction Stoichiometry

• Example Reaction: $2 C<em>8H</em>{18}(l) + 25 O<em>2(g) \rightarrow 16 CO</em>2(g) + 18 H_2O(g)$

  • This equation states that 2 molecules of C8H18 react with 25 molecules of O2 to produce 16 molecules of CO2 and 18 molecules of H2O.

  • In terms of moles:
    $2 \text{ mol } C<em>8H</em>{18} : 25 \text{ mol } O<em>2 : 16 \text{ mol } CO</em>2 : 18 \text{ mol } H_2O$

Mole-to-Mole Conversions

• Conversion Factor: The coefficients in a balanced chemical equation serve as conversion factors between moles of reactants and products.

• From the previous reaction:

  • The stoichiometric ratio can be defined as:
    $2 \text{ moles } C<em>8H</em>{18} : 16 \text{ moles } CO_2$

Example Calculation: Combustion of C8H18

• If we combust 22.0 moles of C8H18, how many moles of CO2 are produced?

  • Using the stoichiometric ratio:
    $2 \text{ moles } C<em>8H</em>{18} : 16 \text{ moles } CO_2$

• Thus, with 22.0 moles of C8H18, we get:
  $\frac{16 \text{ moles } CO<em>2}{2 \text{ moles } C</em>8H<em>{18}} \times 22.0 \text{ moles } C</em>8H<em>{18} = 176 \text{ moles } CO</em>2$

Mass Conversions in Stoichiometry

• To convert mass of reactants (X) to mass of products (Y):

  1. From Grams to Moles: Divide grams by molar mass of X.

  2. Using Mole Ratios: Apply the mole ratio to find the moles of Y.

  3. From Moles to Grams: Multiply moles of Y by its molar mass.

• Graphical representation of conversions:

  • Grams of X -> Moles of X (using molar mass) -> Moles of Y (using mole ratio) -> Grams of Y (using molar mass)

Practice Problem: Photosynthesis Reaction

• In photosynthesis, plants convert CO2 and H2O into glucose (C6H12O6).

• If a plant consumes 37.8g of CO2 in a week, determine the mass of glucose produced.

Example: Antacid Reaction

• Considering magnesium hydroxide (Mg(OH)2) neutralizes stomach acid (HCl):

  • For 3.26 g of Mg(OH)2: how much HCl is neutralized?

  • Calculation:
    3.26 g Mg(OH){2} \times \frac{1 \text{ mol }}{58.35 g Mg(OH){2}} \times \frac{2 \text{ mol HCl}}{1 \text{ mol Mg(OH)_{2}}} \times 36.46 g HCl = 4.07 g HCl

Limiting Reactant Concept

• In reactions with multiple reactants, one reactant may be depleted first:

  • Limiting Reactant: The one that is consumed completely and limits product formation.

  • Reactant in Excess: Any reactant left over after the reaction.

  • Theoretical Yield: The maximum amount of product that can be generated from the limiting reactant.

Example: Combustion of Methane

• Given 5 moles of CH4 and 8 moles of O2:

  • Determine the limiting reactant:

  • Calculation for CO2 production from 5 moles of CH4:

  • For 8 moles of O2, reaction can produce only 4 moles of CO2 (since $2 \text{ O}<em>2 \to 1 \text{ CO}</em>2$).

  • So, O2 is the limiting reactant.

  • Theoretical yield of CO2: 4 moles (from the limiting reactant).

Percent Yield Calculation

• Theoretical Yield: Amount of product expected based on starting amount of limiting reactant.

• Actual Yield: Amount of product actually obtained from the reaction.

• Percent Yield Formula:
  $ext{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$

• Example: For the reaction of 5.00 g of magnesium with silver chloride producing 40.5 g of silver metal:

  • Theoretical yield of silver: 44.4 g

  • Calculation:
    $\text{Percent Yield} = \frac{40.5 g \text{ Ag}}{44.4 g \text{ Ag}} \times 100 = 91.2 \%$

Solutions: Definition and Concentration

• Solution: Homogeneous mixture of solute and solvent.

  • Solute: The component in lesser quantity (changes state).

  • Solvent: The major component (maintains state).

  • An aqueous solution uses water as the solvent. Example: Dissolving table salt in water.

Solutions Composition

• Solutions can be characterized by concentration.

• Dilute: Small amounts of solute relative to solvent.

• Concentrated: Larger amounts of solute relative to solvent.

Molarity (M)

• Definition: Amount of solute (in moles) per liter of solution:
  $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

Preparing Solutions Example

• To prepare a 1.00 M NaCl solution:

  1. Measure 58.44 g of NaCl (1.00 mol).

  2. Dissolve in water until reaching the 1-liter mark.

Practice: Molarity Calculations

• If 25.5 g of KBr is dissolved in enough water to make 1.75 L of solution, calculate molarity: 0.117 M.

Solution Dilution Principles

• Stock solutions of higher concentration are diluted to obtain lower concentrations:

  • Dilution Equation:
    $M<em>1V</em>1 = M<em>2V</em>2$
    where M is molarity and V is volume.

Preparing a Diluted Solution Example

• To prepare 3.00 L of 0.500 M CaCl2 from 10.0 M stock:

  • Measure 0.150 L of the stock.

  • Dilute with water to the total volume of 3.00 L.

Factors Affecting Solubility

1. Temperature: Solubility of solids increases with temperature.

2. Particle Size: Smaller particles dissolve faster.

3. Mixing: Increases dissolution rate.

Types of Solubility and Solutions

• Saturated: Maximum solute dissolved at a given temperature.

• Unsaturated: Less than the maximum solute dissolved.

• Supersaturated: More solute than should theoretically dissolve at the given conditions.

Ion Interactions in Solutions

• Strong electrolytes: Dissolve completely into ions (e.g., NaCl).

• Weak electrolytes: Partially dissolve (e.g., acetic acid).

Dissociation of Ionic Compounds

• Upon dissolution in water, ions separate:

• Example:
  $Na_2S(aq) \rightarrow 2 Na^+(aq) + S^{2-}(aq)$

Solubility Rules

• Soluble Ions: Li+, Na+, K+, NH4+, NO3−, C2H3O2−, etc.

• Insoluble Ions: OH−, S2−, etc. (except with specific ions like Li+, Na+, etc.).

Evidence for Chemical Change

1. Color change

2. Formation of a solid

3. Gas generation

4. Temperature change (heat absorption/release)

Classifications of Chemical Reactions

• Double Replacement: AB + CD → AD + CB

• Single Replacement: A + BC → AC + B

• Decomposition: AB → A + B

• Combination: A + B → AB

• Combustion: CxHy + O2 → CO2 + H2O

Precipitation Reactions

• A solid (precipitate) forms from the mixing of two aqueous solutions, necessitating knowledge of solubility.

Predicting Precipitation Reactions Steps

1. Identify ions present in the reactants.

2. Predict potential products by ion exchange.

3. Determine solubility of each product.

4. Write balanced chemical equation, noting states of matter.

Molecular, Ionic, and Net Ionic Equations

• Molecular Equation: Indicates all compounds in their complete forms.

• Complete Ionic Equation: Shows all particles freely present in solution.

• Net Ionic Equation: Displays only the particles that undergo a chemical change, omitting spectator ions.

Acid-Base Reactions

• Neutralization: Acid reacts with base to yield water and salt.

• Arrhenius Definitions:

  • Acid: Produces H+ ions (e.g., HCl)

  • Base: Produces OH- ions (e.g., NaOH)

• Net ionic equation for neutralization:
  $H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$

Gas-Evolution Reactions

• Gas forms as a result of a chemical reaction, resulting in bubbling.

• E.g.:
  $NaHCO<em>3(aq) + HCl(aq) \rightarrow NaCl(aq) + H</em>2O(l) + CO_2(g)$

Stoichiometry in Solutions

• Stoichiometry with solutions involves using molarity to find relationships between moles of reactions.

Titrations in Acid-Base Chemistry

• Involves mixing an unknown concentration solution with a known solution until the endpoint is reached, typically indicated by a color change with an indicator.

• At the endpoint, moles of acid equal moles of base; stoichiometric calculations may be performed to find concentrations.

Oxidation-Reduction (Redox Reactions)

• Oxidation: Loss of electrons (increase in oxidation state).

• Reduction: Gain of electrons (decrease in oxidation state).

• Agents:

  • Reducing Agent: Contains element that is oxidized.

  • Oxidizing Agent: Contains element that is reduced.

Oxidation States - Rules

1. Free elements = 0.

2. Monatomic ions = charge.

3. Sum of oxidation states in a compound = 0.

4. For polyatomic ions, sum = charge on ion.

5. Group I metals = +1, Group II metals = +2 in compounds.

6. Nonmetals vary (use priority order).

Identifying Redox Reactions

• Ascertain changes in oxidation state to determine whether oxidation and reduction are taking place, as in the example of methane combustion.

  • For instance, carbon undergoes oxidation from (-4) to (+4), while oxygen is reduced from (0) to (-2).