Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
30 g of Mg and 48 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?
In the reaction 2Mg + O2 → 2MgO, 30 g of Mg reacts completely (the other reactants are in excess) and 27.35 g of MgO 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. What volume V1 of stock solution do you need?
250 mL of a 0.25 mol/L solution is mixed with 400 mL of a 2.5 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
What volume V does 470 g of O₂ occupy at 150 kPa and 10 °C? Treat it as an ideal gas.
A fixed amount of Ar goes from 200 kPa, 2 L, 40 °C to 650 kPa and 200 °C. Find the new volume V2.
A 50 g block of lead (c = 0.129 J/(g·°C)) at 190 °C is dropped into 250 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 50 g of C3H8 (M = 44.1 g/mol) burns completely: C3H8 + 5O₂ → 3CO₂ + 4H₂O(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 = 49. Find the equilibrium concentration [HI].
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.1 at the reaction temperature. A container starts with 0.8 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
Find the pH pH of a 0.1 mol/L solution of hypochlorous acid (Ka = 3 × 10⁻⁸).
A buffer contains 0.4 mol/L hydrofluoric acid and 0.5 mol/L of its sodium salt. Ka = 0.00068. Find the pH pH of the buffer.
In a first-order reaction the concentration of A falls from 0.5 mol/L to 0.25 mol/L in 15 min. Find the rate constant k.
Two experiments measure the initial rate of A → products. With CA0 = 0.1 mol/L the rate is 0.01 mol/(L·s); with CA0 = 0.2 mol/L it is 0.02 mol/(L·s). Find the reaction order n with respect to A.
Answer: m = 49.73 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 = 54.99 %
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: V1 = 66.67 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 = 1.635 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: V = 230.5 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: V2 = 0.9298 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Answer: Tf = 18.06 °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 = 2517 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: [HI] = 0.7778 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] = 0.2372 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 = 4.262
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 = 3.264
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: k = 0.04621 min⁻¹
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: n = 1
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.
ExamProblems · examproblems.com/courses/general-chemistry