Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
30 g of Al and 64 g of Cl2 react according to 2Al + 3Cl2 → 2AlCl3. What mass m of AlCl3 can form?
In the reaction 2Al + 3Cl2 → 2AlCl3, 20 g of Al reacts completely (the other reactants are in excess) and 69.19 g of AlCl3 is collected. Find the percent yield Y.
You need 200 mL of 0.1 mol/L solution and have a 0.5 mol/L stock solution. How much water Vwater do you add to the stock solution?
150 mL of a 1 mol/L solution is mixed with 100 mL of a 2.5 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A tank of 5 L contains 430 g of Ar at 30 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of Ar at 1000 kPa and 150 °C, treating it as an ideal gas.
A 75 g block of copper (c = 0.385 J/(g·°C)) at 200 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 22 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find the heat q released when 20 g of C2H5OH (M = 46.07 g/mol) burns completely: C2H5OH + 3O₂ → 2CO₂ + 3H₂O(l). Give q as a positive number.
0.4 mol/L of H₂ and 0.4 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 64. Find the equilibrium concentration [H2].
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.05 at the reaction temperature. A container starts with 1 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
Find the pH pH of a 0.2 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).
A buffer contains 0.5 mol/L acetic acid and 0.3 mol/L of its sodium salt. Ka = 1.8 × 10⁻⁵. Find the pH pH of the buffer.
A first-order reaction in a batch reactor has k = 0.15 min⁻¹. How long does it take to reach 60% conversion? Find t.
A first-order reaction has k = 0.002 min⁻¹ at 20 °C and an activation energy of 80 kJ/mol. The batch reactor is now run at 40 °C for 30 min. Find the conversion X reached.
Answer: m = 80.24 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 = 70 %
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 = 160 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 = 1.6 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 = 5426 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 11.36 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: Tf = 26.79 °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 = 593.3 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: [H2] = 0.08 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] = 0.8 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 = 2.454
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 = 4.523
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: t = 6.109 min
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: X = 38.63 %
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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