Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
20 g of Mg and 8 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?
In the reaction N2 + 3H2 → 2NH3, 20 g of N2 reacts completely (the other reactants are in excess) and 19.94 g of NH3 is collected. Find the percent yield Y.
You need 250 mL of 0.5 mol/L solution and have a 1.5 mol/L stock solution. How much water Vwater do you add to the stock solution?
100 mL of a 1.5 mol/L solution is mixed with 200 mL of a 0.1 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A tank of 5 L contains 210 g of O₂ at 55 °C. Treat it as an ideal gas and find the pressure P.
A fixed amount of CO₂ goes from 150 kPa, 5 L, 10 °C to 650 kPa and 150 °C. Find the new volume V2.
A 250 g block of iron (c = 0.449 J/(g·°C)) at 200 °C is dropped into 400 g of water (c = 4.18 J/(g·°C)) at 23 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find ΔH° for the reaction CH4 + 2O₂ → CO₂ + 2H₂O(l).
0.25 mol/L of H₂ and 0.25 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 4. Find the equilibrium concentration [HI].
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.05 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 0.2 mol/L solution of acetic acid (Ka = 1.8 × 10⁻⁵).
A 1.00 L buffer contains 0.2 mol formic acid and 0.2 mol of its conjugate base (Ka = 0.00018). 0.025 mol of NaOH is added. Find the new pH pH.
A first-order reaction in a batch reactor has k = 0.1 min⁻¹. How long does it take to reach 95% conversion? Find t.
Two experiments measure the initial rate of A → products. With CA0 = 0.5 mol/L the rate is 0.01 mol/(L·s); with CA0 = 2 mol/L it is 0.64 mol/(L·s). Find the reaction order n with respect to A.
Answer: m = 20.15 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 = 82.02 %
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 = 166.7 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 = 0.5667 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 = 3581 kPa
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 1.724 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Tf = 34.14 °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° = −890.3 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.
Negative: the reaction gives off heat (exothermic).
Answer: [HI] = 0.25 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] = 0.1351 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 = 2.724
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 = 3.854
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 = 29.96 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: n = 3
Method of initial rates: for r = k·Cⁿ, comparing two runs that differ only in the starting concentration removes k: r₂/r₁ = (C₂/C₁)ⁿ. Doubling C doubles the rate for first order and quadruples it for second order.
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