General Chemistry Exam 2 Review: Stoichiometry, Reactions, and Solution Chemistry

Stoichiometry and Chemical Reactions

Balancing Chemical Equations

Balancing chemical equations ensures the law of conservation of mass is obeyed. Each side of the equation must have the same number of atoms for each element.

• Key Point: Adjust coefficients (not subscripts) to balance atoms.

• Example: $\mathrm{H_2SO_4 (aq) + 2LiOH (aq) \rightarrow Li_2SO_4 (aq) + 2H_2O (l)}$

Types of Chemical Reactions

Chemical reactions can be classified into several types, including combination, decomposition, single displacement, double displacement, and combustion reactions.

• Combination Reaction: Two or more substances combine to form one product.

• Decomposition Reaction: One substance breaks down into two or more products.

• Example: $\mathrm{2NaN_3 (s) \rightarrow 2Na (s) + 3N_2 (g)}$ (decomposition of sodium azide)

Stoichiometric Calculations

Percent Composition

Percent composition by mass expresses the mass percentage of each element in a compound.

• Formula: $\%\,\text{element} = \frac{\text{mass of element in 1 mol of compound}}{\text{molar mass of compound}} \times 100\%$

• Example: Calculate the percent of sulfur in $\mathrm{Al_2(SO_4)_3}$.

Empirical and Molecular Formulas

The empirical formula gives the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms of each element in a molecule.

• Steps to Determine Empirical Formula:

  1. Convert mass percentages to grams (assume 100 g sample).

  2. Convert grams to moles for each element.

  3. Divide by the smallest number of moles to get the simplest ratio.

  4. Multiply to obtain whole numbers if necessary.

• Example: A compound contains 47.5% C, 10.6% H, and 41.9% N by mass. Find the empirical formula.

Limiting Reactant and Theoretical Yield

The limiting reactant is the reactant that is completely consumed first, limiting the amount of product formed. The theoretical yield is the maximum amount of product that can be formed from the limiting reactant.

• Steps:

  1. Convert all reactant quantities to moles.

  2. Use stoichiometry to determine which reactant produces the least amount of product.

  3. The reactant that produces the least is the limiting reactant.

  4. Calculate the theoretical yield based on the limiting reactant.

• Example: $\mathrm{Fe_2O_3 (s) + 3CO (g) \rightarrow 2Fe (s) + 3CO_2 (g)}$

Percent Yield

Percent yield compares the actual yield to the theoretical yield.

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

Solution Chemistry

Molarity and Dilution

Molarity (M) is the number of moles of solute per liter of solution. Dilution involves adding solvent to decrease the concentration of a solution.

• Formula: $M_1V_1 = M_2V_2$

• Example: If 11.1 M HNO3 is diluted from 250.0 mL to 500.0 mL, the new concentration is $M_2 = \frac{M_1V_1}{V_2}$.

Precipitation Reactions and Solubility

Precipitation reactions occur when two aqueous solutions combine to form an insoluble solid (precipitate). Solubility rules help predict the formation of precipitates.

• Example: Mixing K2SO4 and Ba(NO3)2 forms BaSO4 (s) as a precipitate.

Electrolytes and Acids/Bases

Electrolytes are substances that conduct electricity in aqueous solution. Strong acids and bases dissociate completely, while weak acids and bases only partially dissociate.

• Strong Acids: HCl, HNO3, H2SO4

• Strong Bases: NaOH, KOH

• Example: Acetic acid (CH3COOH) is a weak acid; HCl is a strong acid.

Ionic and Net Ionic Equations

Ionic equations show all ions present in a reaction. Net ionic equations show only the species that actually change during the reaction.

• Example: $\mathrm{HCl (aq) + NaOH (aq) \rightarrow NaCl (aq) + H_2O (l)}$

• Net Ionic Equation: $\mathrm{H^+ (aq) + OH^- (aq) \rightarrow H_2O (l)}$

Redox Reactions and Oxidation Numbers

Assigning Oxidation Numbers

Oxidation numbers are assigned to atoms to keep track of electron transfer in redox reactions.

• Rules:

  1. Elements in their standard state have oxidation number 0.

  2. Group 1 metals: +1; Group 2 metals: +2; Fluorine: -1; Oxygen: usually -2; Hydrogen: +1 with nonmetals, -1 with metals.

  3. The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion, it equals the ion's charge.

• Example: In KMnO4, Mn has an oxidation number of +7.

Identifying Oxidation and Reduction

Oxidation is the loss of electrons (increase in oxidation number), and reduction is the gain of electrons (decrease in oxidation number).

• Example: In $\mathrm{Cr_2O_7^{2-} + 2NO (g) \rightarrow 2NO_3^- (aq) + 2Cr^{3+} (aq)}$, Cr is reduced and N is oxidized.

Gas Laws and Stoichiometry

Gas Stoichiometry

Gas stoichiometry involves calculations using the relationships between moles, volume, and molar mass for gases, often at standard temperature and pressure (STP).

• Key Point: 1 mole of an ideal gas occupies 22.4 L at STP.

• Example: Calculate the mass of NaN3 needed to produce a given volume of N2 gas.

Practice Problems and Applications

• Calculate the number of atoms or molecules in a given mass using Avogadro's number ($6.022 \times 10^{23}$).

• Determine the empirical formula from percent composition data.

• Identify the limiting reactant in a chemical reaction and calculate the theoretical yield.

• Predict the formation of a precipitate using solubility rules.

• Write net ionic equations for acid-base and precipitation reactions.

• Assign oxidation numbers and identify redox changes in reactions.

Summary Table: Key Concepts in Stoichiometry and Reactions

Concept	Definition	Key Formula/Rule	Example
Percent Composition	Mass % of each element in a compound	$\% = \frac{\text{mass of element}}{\text{molar mass}} \times 100$	S in Al2(SO4)3
Empirical Formula	Simplest whole-number ratio of atoms	Convert % to g, then to mol, divide by smallest	C2H6O
Limiting Reactant	Reactant used up first	Compare mole ratios from balanced equation	CO in Fe2O3 + 3CO
Molarity	Moles of solute per liter of solution	$M = \frac{\text{mol}}{\text{L}}$	11.1 M HNO3
Dilution	Decrease concentration by adding solvent	$M_1V_1 = M_2V_2$	250 mL to 500 mL dilution
Net Ionic Equation	Shows only reacting ions	Remove spectator ions	$\mathrm{H^+ + OH^- \rightarrow H_2O}$
Oxidation Number	Charge assigned to atom	Rules for assigning values	Mn in KMnO4: +7

Additional info: Some steps and explanations have been expanded for clarity and completeness, including definitions, formulas, and examples relevant to General Chemistry topics such as stoichiometry, solution chemistry, and redox reactions.