Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
16 g of Al and 16 g of Cl2 react according to 2Al + 3Cl2 → 2AlCl3. What mass m of AlCl3 can form?
In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 8 g of Fe2O3 reacts completely (the other reactants are in excess) and 5.42 g of CO2 is collected. Find the percent yield Y.
You need 250 mL of 0.3 mol/L solution and have a 3 mol/L stock solution. What volume V1 of stock solution do you need?
200 mL of a 1 mol/L solution is mixed with 250 mL of a 3 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A tank of 30 L contains 310 g of N₂ at 15 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of Ar goes from 300 kPa, 10 L, 15 °C to 150 kPa and 250 °C. Find the new volume V2.
A 75 g block of aluminium (c = 0.897 J/(g·°C)) at 140 °C is dropped into 100 g of water (c = 4.18 J/(g·°C)) at 25 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find the heat q released when 100 g of C4H10 (M = 58.12 g/mol) burns completely: 2C4H10 + 13O₂ → 8CO₂ + 10H₂O(l). Give q as a positive number.
0.5 mol/L of H₂ and 0.5 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 54. Find the equilibrium concentration [HI].
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.05 at the reaction temperature. A container starts with 0.2 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
Find the pH pH of a 0.05 mol/L solution of formic acid (Ka = 0.00018).
A 1.00 L buffer contains 0.1 mol hypochlorous acid and 0.1 mol of its conjugate base (Ka = 3 × 10⁻⁸). 0.05 mol of NaOH is added. Find the new pH pH.
A reaction with rate law −rA = k·CA² has k = 0.8 L/(mol·min) and starts at 2 mol/L. Find the concentration CA after 2 min.
The rate constant of a reaction is 0.002 min⁻¹ at 30 °C and 0.0712 min⁻¹ at 60 °C. Find the activation energy Ea.
Answer: m = 20.06 g
Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.
Answer: Y = 81.94 %
Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.
Answer: V1 = 25 mL
Molarity: concentration c = n/V in mol/L, with n = m/M. Diluting adds solvent but not solute, so the number of moles stays the same: c₁·V₁ = c₂·V₂. When two solutions of the same solute are mixed, the moles add up and so do the volumes.
Answer: c = 2.111 mol/L
Molarity: concentration c = n/V in mol/L, with n = m/M. Diluting adds solvent but not solute, so the number of moles stays the same: c₁·V₁ = c₂·V₂. When two solutions of the same solute are mixed, the moles add up and so do the volumes.
Answer: P = 883.8 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 36.31 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Tf = 40.94 °C
Heat and enthalpy: heating a substance takes q = m·c·ΔT. In a calorimeter the heat lost by the hot object equals the heat gained by the cold one. The enthalpy change of a reaction follows from enthalpies of formation: ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants), each multiplied by its coefficient.
Answer: q = 4951 kJ
Heat and enthalpy: heating a substance takes q = m·c·ΔT. In a calorimeter the heat lost by the hot object equals the heat gained by the cold one. The enthalpy change of a reaction follows from enthalpies of formation: ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants), each multiplied by its coefficient.
Answer: [HI] = 0.7861 mol/L
Chemical equilibrium: for aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]c[D]d/([A]a[B]b), using concentrations at equilibrium. An ICE table (Initial, Change, Equilibrium) links the starting amounts to the equilibrium amounts through one unknown x, scaled by the coefficients.
Answer: [PCl5] = 0.1219 mol/L
Chemical equilibrium: for aA + bB ⇌ cC + dD, the equilibrium constant is Kc = [C]c[D]d/([A]a[B]b), using concentrations at equilibrium. An ICE table (Initial, Change, Equilibrium) links the starting amounts to the equilibrium amounts through one unknown x, scaled by the coefficients.
Answer: pH = 2.536
pH: pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻], with pH + pOH = 14 at 25 °C. Strong acids and bases ionise completely. A weak acid only partly: Ka = x²/(c − x), where x = [H₃O⁺]. A buffer of a weak acid and its conjugate base follows Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]).
HA ⇌ H⁺ + A⁻: x mol/L of acid ionises.
Answer: pH = 8
pH: pH = −log₁₀[H₃O⁺] and pOH = −log₁₀[OH⁻], with pH + pOH = 14 at 25 °C. Strong acids and bases ionise completely. A weak acid only partly: Ka = x²/(c − x), where x = [H₃O⁺]. A buffer of a weak acid and its conjugate base follows Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]).
Answer: CA = 0.4762 mol/L
Reaction rates: for a first-order reaction −dC/dt = kC, so C = C₀·e−kt and the half-life ln 2/k does not depend on concentration. For a second-order reaction 1/C = 1/C₀ + kt. The rate constant rises with temperature following Arrhenius: k = A·e−Ea/(RT).
Answer: Ea = 99.99 kJ/mol
Reaction rates: for a first-order reaction −dC/dt = kC, so C = C₀·e−kt and the half-life ln 2/k does not depend on concentration. For a second-order reaction 1/C = 1/C₀ + kt. The rate constant rises with temperature following Arrhenius: k = A·e−Ea/(RT).
ExamProblems · examproblems.com/courses/general-chemistry