Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
24 g of Al and 8 g of Cl2 react according to 2Al + 3Cl2 → 2AlCl3. What mass m of AlCl3 can form?
In the reaction C6H12O6 → 2C2H5OH + 2CO2, 30 g of C6H12O6 reacts completely (the other reactants are in excess) and 9.51 g of C2H5OH is collected. Find the percent yield Y.
You need 250 mL of 0.4 mol/L solution and have a 1.5 mol/L stock solution. How much water Vwater do you add to the stock solution?
300 mL of a 0.25 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 50 L contains 270 g of N₂ at 5 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of He goes from 300 kPa, 10 L, 25 °C to 800 kPa and 110 °C. Find the new volume V2.
A 250 g block of lead (c = 0.129 J/(g·°C)) at 120 °C is dropped into 300 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 the heat q released when 250 g of C6H12O6 (M = 180.16 g/mol) burns completely: C6H12O6 + 6O₂ → 6CO₂ + 6H₂O(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 = 9. Find the equilibrium concentration [H2].
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.02 at the reaction temperature. A container starts with 0.5 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
Find the pH pH of a 0.1 mol/L solution of acetic acid (Ka = 1.8 × 10⁻⁵).
A 1.00 L buffer contains 0.1 mol formic acid and 0.15 mol of its conjugate base (Ka = 0.00018). 0.02 mol of NaOH is added. Find the new pH pH.
A second-order reaction (−rA = k·CA²) in a batch reactor has k = 0.25 L/(mol·min) and starts at 2 mol/L. How long does it take to reach 80% conversion? Find t.
The rate constant of a reaction is 0.02 min⁻¹ at 40 °C and 0.201 min⁻¹ at 60 °C. Find the activation energy Ea.
Answer: m = 10.03 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 = 61.98 %
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 = 183.3 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 = 0.8125 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 = 445.8 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 4.819 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Tf = 19.58 °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 = 3888 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] = 0.16 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] = 0.0905 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 = 2.875
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 = 4.072
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 = 8 min
Integrated rate laws: zero order: C = C₀ − kt (straight line, t½ = C₀/(2k)); first order: ln(C₀/C) = kt (t½ = ln 2/k); second order: 1/C − 1/C₀ = kt (t½ = 1/(kC₀)). Conversion X means C = C₀(1 − X).
Answer: Ea = 100.1 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).
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