Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
20 g of C3H8 and 8 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of H2O can form?
In the reaction N2 + 3H2 → 2NH3, 8 g of N2 reacts completely (the other reactants are in excess) and 7.29 g of NH3 is collected. Find the percent yield Y.
You need 200 mL of 0.05 mol/L solution and have a 1.5 mol/L stock solution. How much water Vwater do you add to the stock solution?
50 mL of a 0.1 mol/L solution is mixed with 400 mL of a 0.3 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
What volume V does 500 g of He occupy at 700 kPa and 25 °C? Treat it as an ideal gas.
Find the density ρ of N₂ at 400 kPa and 50 °C, treating it as an ideal gas.
A 75 g block of copper (c = 0.385 J/(g·°C)) at 80 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 19 °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 C4H10 (M = 58.12 g/mol) burns completely: 2C4H10 + 13O₂ → 8CO₂ + 10H₂O(l). Give q as a positive number.
1 mol/L of H₂ and 1 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 50. Find the equilibrium concentration [HI].
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.04 at the reaction temperature. A container starts with 0.1 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
Find the pH pH of a 0.25 mol/L solution of acetic acid (Ka = 1.8 × 10⁻⁵).
A 1.00 L buffer contains 0.1 mol formic acid and 0.1 mol of its conjugate base (Ka = 0.00018). 0.02 mol of NaOH is added. Find the new pH pH.
In a first-order reaction the concentration of A falls from 1.2 mol/L to 0.96 mol/L in 40 min. Find the rate constant k.
Two experiments measure the initial rate of A → products. With CA0 = 0.4 mol/L the rate is 0.008 mol/(L·s); with CA0 = 0.8 mol/L it is 0.0226 mol/(L·s). Find the reaction order n with respect to A.
Answer: m = 3.604 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 = 74.97 %
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 = 193.3 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 = 0.2778 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 = 442.3 L
Ideal gas law: PV = nRT, with n = m/M.
Answer: ρ = 4.17 kg/m³
Ideal gas density: ρ = PM/(RT).
Answer: Tf = 20.64 °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 = 4951 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] = 1.559 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] = 0.05367 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 = 3.921
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 = 0.005579 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 = 1.498
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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