Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
30 g of Al and 16 g of Cl2 react according to 2Al + 3Cl2 → 2AlCl3. What mass m of AlCl3 can form?
In the reaction 2H2 + O2 → 2H2O, 20 g of H2 reacts completely (the other reactants are in excess) and 139.4 g of H2O is collected. Find the percent yield Y.
You need 100 mL of 0.25 mol/L solution and have a 3 mol/L stock solution. How much water Vwater do you add to the stock solution?
50 mL of a 0.8 mol/L solution is mixed with 400 mL of a 1.2 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A tank of 50 L contains 80 g of CH₄ at 90 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of CO₂ at 400 kPa and 150 °C, treating it as an ideal gas.
A 50 g block of zinc (c = 0.388 J/(g·°C)) at 190 °C is dropped into 200 g of water (c = 4.18 J/(g·°C)) at 17 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find ΔH° for the reaction C2H5OH + 3O₂ → 2CO₂ + 3H₂O(l).
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 = 9. Find the equilibrium concentration [HI].
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 at the reaction temperature. A container starts with 0.1 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
Find the pH pH of a 0.25 mol/L solution of formic acid (Ka = 0.00018).
A 1.00 L buffer contains 0.1 mol benzoic acid and 0.25 mol of its conjugate base (Ka = 6.3 × 10⁻⁵). 0.02 mol of NaOH is added. Find the new pH pH.
A reaction with rate law −rA = k·CA² has k = 0.1 L/(mol·min) and starts at 0.5 mol/L. Find the concentration CA after 10 min.
A reaction has an activation energy of 80 kJ/mol. At 50 °C it is too slow. To what temperature T2 (in °C) must it be heated to make it 4 times faster?
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 = 78 %
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: Vwater = 91.67 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 = 1.156 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 = 301.2 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 5.004 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: Tf = 20.92 °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: ΔH° = −1367 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.
Negative: the reaction gives off heat (exothermic).
Answer: [HI] = 0.6 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] = 0.04202 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 = 2.179
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 = 4.729
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 = 0.3333 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: T2 = 65.78 °C
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).
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