Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
30 g of H2 and 16 g of O2 react according to 2H2 + O2 → 2H2O. What mass m of H2O can form?
In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 12 g of Fe2O3 reacts completely (the other reactants are in excess) and 6.15 g of CO2 is collected. Find the percent yield Y.
You need 100 mL of 0.2 mol/L solution and have a 2 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 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 5 L contains 120 g of CO₂ at 50 °C. Treat it as an ideal gas and find the pressure P.
Find the density ρ of CO₂ at 150 kPa and 50 °C, treating it as an ideal gas.
A 75 g block of iron (c = 0.449 J/(g·°C)) at 170 °C is dropped into 400 g of water (c = 4.18 J/(g·°C)) at 24 °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 C6H12O6 (M = 180.16 g/mol) burns completely: C6H12O6 + 6O₂ → 6CO₂ + 6H₂O(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 = 25. Find the equilibrium concentration [H2].
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.01 at the reaction temperature. A container starts with 0.2 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
Find the pH pH of a 0.25 mol/L solution of acetic acid (Ka = 1.8 × 10⁻⁵).
A buffer contains 0.4 mol/L acetic acid and 0.5 mol/L of its sodium salt. Ka = 1.8 × 10⁻⁵. Find the pH pH of the buffer.
A reaction with rate law −rA = k·CA² has k = 0.05 L/(mol·min) and starts at 2 mol/L. Find the concentration CA after 15 min.
A reaction has an activation energy of 70 kJ/mol. At 30 °C it is too slow. To what temperature T2 (in °C) must it be heated to make it 10 times faster?
Answer: m = 18.02 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 = 61.99 %
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 = 90 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 = 2.633 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 = 1465 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 2.457 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: Tf = 26.88 °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 = 1555 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] = 0.1429 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: [Cl2] = 0.04 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 = 2.675
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 = 4.842
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 = 0.8 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 = 57.4 °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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