24 g of Mg and 40 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?
M(Mg) = 24.31M(O2) = 32M(MgO) = 40.3 g/mol
m = g
2.StoichiometryHard
In the reaction CH4 + 2O2 → CO2 + 2H2O, 8 g of CH4 reacts completely (the other reactants are in excess) and 12.07 g of CO2 is collected. Find the percent yield Y.
M(CH4) = 16.04 g/molM(CO2) = 44.01 g/molactual: 12.07 g
Y = %
3.Molarity & dilutionMedium
You need 500 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?
c1 = 1.5 mol/Lc2 = 0.4 mol/LV2 = 500 mL
V1 = mL
4.Molarity & dilutionHard
200 mL of a 1.5 mol/L solution is mixed with 400 mL of a 0.05 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 75 g block of copper (c = 0.385 J/(g·°C)) at 120 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 16 °C in an insulated cup. Find the final temperature Tf.
0.8 mol/L of H₂ and 0.8 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 9. Find the equilibrium concentration [HI].
[H₂]₀ = [I₂]₀ = 0.8 mol/LKc = 9
[HI] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.01 at the reaction temperature. A container starts with 1 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
[PCl₅]₀ = 1 mol/LKc = 0.01
[PCl5] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.1 mol/L solution of acetic acid (Ka = 1.8 × 10⁻⁵).
c = 0.1 mol/LKa = 1.8 × 10⁻⁵
pH =
12.pH & buffersHard
A buffer contains 0.4 mol/L hydrofluoric acid and 0.05 mol/L of its sodium salt. Ka = 0.00068. Find the pH pH of the buffer.
[HA] = 0.4[A⁻] = 0.05Ka = 0.00068
pH =
13.Reaction kineticsMedium
A reaction with rate law −rA = k·CA² has k = 1 L/(mol·min) and starts at 0.5 mol/L. Find the concentration CA after 20 min.
k = 1 L/(mol·min)CA0 = 0.5 mol/Lt = 20 min
CA = mol/L
14.Reaction kineticsHard
The rate constant of a reaction is 0.002 min⁻¹ at 35 °C and 0.00711 min⁻¹ at 50 °C. Find the activation energy Ea.
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.
Moles of each reactant
n(Mg) = 24/24.31 = 0.9872 mol
n(O2) = 40/32 = 1.25 mol
Limiting reactant
0.9872/2 = 0.4936, 1.25/1 = 1.25 → Mg
Mass of product
n(MgO) = 0.4936·2 = 0.9872 mol
m = 0.9872·40.3 = 39.79 g
Answer
m = 39.79 g
2.Stoichiometry
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.
Theoretical yield
n(CH4) = 8/16.04 = 0.4988 mol
m = 0.4988·1/1·44.01 = 21.95 g
Percent yield
Y = 12.07/21.95·100 = 54.99%
Answer
Y = 54.99%
3.Molarity & dilution
Answer: V1 = 133.3 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.
Volume of stock solution
V1 = c2·V2/c1 = 0.4·500/1.5 = 133.3 mL
Answer
V1 = 133.3 mL
4.Molarity & dilution
Answer: c = 0.5333 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.
Total amount
n = 1.5·200 + 0.05·400 = 320 mmol
Concentration of the mixture
c = 320/600 = 0.5333 mol/L
Answer
c = 0.5333 mol/L
5.Ideal gas law
Answer: V = 1085 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 420/4.003 = 104.9 mol
T = 100 + 273.15 = 373.1 K
Volume
V = nRT/P = 104.9·8.314·373.1/300 = 1085 L
Answer
V = 1085 L
6.Ideal gas law
Answer: ρ = 9.864 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 0 + 273.15 = 273.1 K
Density
ρ = 700·32/(8.314·273.1) = 9.864 kg/m³
Answer
ρ = 9.864 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 18.8 °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.
Heat balance
75·0.385·(120 − Tf) = 250·4.18·(Tf − 16)
Solve for the final temperature
Tf = (28.88·120 + 1045·16)/(28.88 + 1045) = 18.8 °C
Answer
Tf = 18.8 °C
8.Calorimetry & enthalpy
Answer: ΔH° = −1367 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).
9.Chemical equilibrium
Answer: [HI] = 0.96 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.
Equilibrium expression
9 = (2x)²/(0.8 − x)²
Take the square root
3 = 2x/(0.8 − x) → x = 0.8·3/(2 + 3) = 0.48 mol/L
Equilibrium concentration
[HI] = 2x = 0.96 mol/L
Answer
[HI] = 0.96 mol/L
10.Chemical equilibrium
Answer: [PCl5] = 0.9049 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.
Equilibrium expression
0.01 = x²/(1 − x) → x² + 0.01·x − 0.01 = 0
Solve the quadratic
x = (−0.01 + √(0.01² + 4·0.01))/2 = 0.09512 mol/L
Equilibrium concentration
[PCl₅] = 1 − 0.09512 = 0.9049 mol/L
Answer
[PCl5] = 0.9049 mol/L
11.pH & buffers
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]).
Equilibrium
HA ⇌ H⁺ + A⁻: x mol/L of acid ionises.
Ka = x²/(0.1 − x) → x = 0.001333 mol/L
pH
pH = −log(0.001333) = 2.875
Answer
pH = 2.875
12.pH & buffers
Answer: pH = 2.264
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]).
pKa
pKa = 3.167
Henderson–Hasselbalch
pH = 3.167 + log(0.05/0.4) = 2.264
Answer
pH = 2.264
13.Reaction kinetics
Answer: CA = 0.04545 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).
Integrated second-order law
1/CA = 1/CA0 + kt = 2 + 1·20 = 22 L/mol
CA = 1/22 = 0.04545 mol/L
Answer
CA = 0.04545 mol/L
14.Reaction kinetics
Answer: Ea = 70 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).
Absolute temperatures
T1 = 308.1 K, T2 = 323.1 K
Solve Arrhenius for Ea
Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·1.268/(-1.506e-4) = 70004 J/mol