General Chemistry: Chemical Reactions, Solutions, and Gases – Exam 2 Review Notes

Chapter 4: Chemical Reactions and Chemical Quantities

Balancing Chemical Equations

Balancing chemical equations ensures the law of conservation of mass is obeyed, with equal numbers of each atom on both sides of the reaction.

• Key Point: Start by balancing elements that appear in only one reactant and one product first, leaving oxygen and hydrogen for last.

• Example: Formation of nitric oxide and water from ammonia and oxygen:

$4\mathrm{NH}_3 + 5\mathrm{O}_2 \rightarrow 4\mathrm{NO} + 6\mathrm{H}_2\mathrm{O}$

• Tip: Use fractional coefficients if needed, then multiply through to obtain whole numbers.

Stoichiometry and Limiting Reactants

Stoichiometry involves calculating the quantities of reactants and products in a chemical reaction. The limiting reactant is the substance that is completely consumed first, limiting the amount of product formed.

• Key Point: Use molar ratios from the balanced equation to relate amounts of reactants and products.

• Example: Reaction of Ba(NO3)2 and Na2SO4:

$\mathrm{Ba(NO}_3)_2 (aq) + \mathrm{Na}_2\mathrm{SO}_4 (aq) \rightarrow \mathrm{BaSO}_4 (s) + 2\mathrm{NaNO}_3 (aq)$

• Calculate moles using $\text{mol} = M \times V$.

• Find the mass of product using $\text{mass} = \text{mol} \times \text{molar mass}$.

• Percent yield: $\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

• Example Calculation: If 548 g BaSO4 is expected but only 101 g is obtained, percent yield is $\frac{101}{548} \times 100\% = 18\%$.

Limiting Reagent Determination

• Key Point: Convert all reactant quantities to moles, then use stoichiometry to determine which reactant will run out first.

• Example: For 110 g Al and 200 g MnO in $2\mathrm{Al} + 3\mathrm{MnO} \rightarrow \mathrm{Al}_2\mathrm{O}_3 + 3\mathrm{Mn}$:

• Calculate moles: $110\,\mathrm{g} \div 26.98\,\mathrm{g/mol} = 4.1\,\mathrm{mol}$ Al; $200\,\mathrm{g} \div 70.9\,\mathrm{g/mol} = 2.8\,\mathrm{mol}$ MnO.

• Since 3 mol MnO are needed per 2 mol Al, MnO is limiting.

Combustion Reactions

• Definition: Combustion is a reaction with O2 producing CO2 and H2O (for hydrocarbons).

• Example: Propane combustion:

$\mathrm{C}_3\mathrm{H}_8 + 5\mathrm{O}_2 \rightarrow 3\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O}$

Chapter 5: Introduction to Solutions and Aqueous Solutions

Calculating Molarity

Molarity (M) is the number of moles of solute per liter of solution.

• Formula: $M = \frac{\text{mol}}{\text{L}}$

• Example: 10.1 g NaCl in 1.52 L:

• Convert grams to moles: $10.1\,\mathrm{g} \div 58.44\,\mathrm{g/mol} = 0.173\,\mathrm{mol}$

• Molarity: $0.173\,\mathrm{mol} \div 1.52\,\mathrm{L} = 0.114\,\mathrm{M}$

• Note: The original calculation used 59.44 g/mol, but the correct molar mass of NaCl is 58.44 g/mol.

Dilution Calculations

• Formula: $C_1V_1 = C_2V_2$

• Example: To make 308 mL of 0.151 M HCl from 0.39 M HCl:

• $V_1 = \frac{C_2V_2}{C_1} = \frac{0.151 \times 0.308}{0.39} = 0.119\,\mathrm{L}$

Precipitation Reactions

• Key Point: A precipitate forms when an insoluble ionic compound is produced in solution.

• Example: Mixing Na2CrO4 and FeCl3 forms Fe2(CrO4)3 as a precipitate.

Molecular, Total Ionic, and Net Ionic Equations

• Molecular Equation: Shows all reactants and products as compounds.

• Total Ionic Equation: Shows all strong electrolytes as ions.

• Net Ionic Equation: Shows only the species that actually change during the reaction.

• Example: CaCl2 + Na2CO3 reaction:

Equation Type	Equation
Molecular	CaCl2 (aq) + Na2CO3 (aq) → CaCO3 (s) + 2NaCl (aq)
Total Ionic	Ca2+ (aq) + 2Cl- (aq) + 2Na+ (aq) + CO32- (aq) → CaCO3 (s) + 2Na+ (aq) + 2Cl- (aq)
Net Ionic	Ca2+ (aq) + CO32- (aq) → CaCO3 (s)

Redox Reactions and Activity Series

• Key Point: A redox reaction occurs if a more active metal displaces a less active metal from solution.

• Example: Iron and zinc oxide: No reaction, as iron is less active than zinc.

• Tin and copper nitrate: Reaction occurs, as tin is more active than copper.

Titration and Acid-Base Reactions

• Definition: Titration is a technique to determine the concentration of a solution by reacting it with a standard solution.

• Formula: $M_1V_1 = M_2V_2$ (for 1:1 stoichiometry)

• Example: 50.00 mL HBr titrated with 42.21 mL of 0.1025 M NaOH:

• Find moles NaOH: $0.04221\,\mathrm{L} \times 0.1025\,\mathrm{mol/L} = 0.004329\,\mathrm{mol}$

• Since 1:1 ratio, moles HBr = moles NaOH

• Molarity HBr: $0.004329\,\mathrm{mol} \div 0.05000\,\mathrm{L} = 0.0866\,\mathrm{M}$

• This is an acid-base neutralization reaction.

Generating Gases from Solutions

• Key Point: Certain gases can be generated by adding specific salts to acids or bases.

• Table: Methods to Generate Gases

Gas	Reagent to Add
CO2	NaHCO3 or Na2CO3 to HCl
H2S	NaHS or Na2S to HCl
SO2	NaHSO3 or Na2SO3 to HCl
NH3	NaOH to NH4Cl solution

Chapter 6: Gases (Sections 6.2–6.5)

Gas Laws

• Boyle's Law: $P_1V_1 = P_2V_2$ (at constant T and n)

• Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ (at constant P and n)

• Combined Gas Law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

• Ideal Gas Law: $PV = nRT$

Example Calculations

• Pressure-Volume Change: If a gas at 1.2 atm and 1.7 L expands to 2.3 L, final pressure is:

$P_2 = \frac{P_1V_1}{V_2} = \frac{1.2 \times 1.7}{2.3} = 0.89\,\mathrm{atm}$

• Temperature-Volume Change: 5.8 L at 92°C (365 K) cooled to 77°C (350 K):

$V_2 = V_1 \times \frac{T_2}{T_1} = 5.8 \times \frac{350}{365} = 5.6\,\mathrm{L}$

• Pressure-Temperature Change: Tire at 40 psi and 80°F (300 K) cooled to 12°F (262 K):

$P_2 = P_1 \times \frac{T_2}{T_1} = 40 \times \frac{262}{300} = 35\,\mathrm{psi}$

• Volume Change in Reaction: $2\mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}$ (all gases):

• Initial moles: 2 + 1 = 3; Final moles: 2. Volume ratio = 2:3.

Calculating Gas Volume and Moles

• Key Point: Use the ideal gas law to relate pressure, volume, temperature, and moles.

• Example: 0.33 mol at 298 K and 1.32 atm:

$V = \frac{nRT}{P} = \frac{0.33 \times 0.0821 \times 298}{1.32} = 6.11\,\mathrm{L}$

Collecting Gases Over Water

• Key Point: When collecting a gas over water, subtract the vapor pressure of water from the total pressure to find the pressure of the dry gas.

• Example: Acetylene collected at 732 mm Hg, water vapor pressure 18 mm Hg:

$P_{\mathrm{C_2H_2}} = 732 - 18 = 714\,\mathrm{mm\ Hg}$

$\text{Convert to atm:}\ 714\,\mathrm{mm\ Hg} \div 760 = 0.939\,\mathrm{atm}$

$\text{Moles:}\ n = \frac{PV}{RT} = \frac{0.939 \times 0.523}{0.0821 \times 293} = 0.0209\,\mathrm{mol}$

Molar Mass of a Volatile Liquid (Dumas Method)

• Key Point: The molar mass of a volatile liquid can be determined by vaporizing a known mass in a known volume at a known temperature and pressure.

• Formula: $\text{Molar mass} = \frac{dRT}{P}$, where $d$ is density ($\text{g/L}$).

• Example: 1.18 g vapor in 0.240 L at 150°C (423 K), 783 mm Hg (1.03 atm):

$\text{Molar mass} = \frac{(1.18/0.240) \times 0.0821 \times 423}{1.03} = 166\,\mathrm{g/mol}$

Additional info: All calculations assume ideal behavior unless otherwise specified. For more complex or real-world scenarios, corrections may be needed.