24 g of C3H8 and 64 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of CO2 can form?
M(C3H8) = 44.1M(O2) = 32M(CO2) = 44.01 g/mol
m = g
2.StoichiometryHard
In the reaction CaCO3 → CaO + CO2, 8 g of CaCO3 reacts completely (the other reactants are in excess) and 3.5 g of CaO is collected. Find the percent yield Y.
M(CaCO3) = 100.09 g/molM(CaO) = 56.08 g/molactual: 3.5 g
Y = %
3.Molarity & dilutionMedium
You need 500 mL of 0.5 mol/L solution and have a 5 mol/L stock solution. How much water Vwater do you add to the stock solution?
c1 = 5 mol/Lc2 = 0.5 mol/LV2 = 500 mL
Vwater = mL
4.Molarity & dilutionHard
100 mL of a 1.5 mol/L solution is mixed with 400 mL of a 0.1 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 150 g block of aluminium (c = 0.897 J/(g·°C)) at 170 °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.
0.2 mol/L of H₂ and 0.2 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 49. Find the equilibrium concentration [HI].
[H₂]₀ = [I₂]₀ = 0.2 mol/LKc = 49
[HI] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 at the reaction temperature. A container starts with 1 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 1 mol/LKc = 0.08
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.05 mol/L solution of acetic acid (Ka = 1.8 × 10⁻⁵).
c = 0.05 mol/LKa = 1.8 × 10⁻⁵
pH =
12.pH & buffersHard
A buffer contains 0.1 mol/L hydrofluoric acid and 0.5 mol/L of its sodium salt. Ka = 0.00068. Find the pH pH of the buffer.
[HA] = 0.1[A⁻] = 0.5Ka = 0.00068
pH =
13.Reaction kineticsMedium
A reaction with rate law −rA = k·CA² has k = 0.5 L/(mol·min) and starts at 2 mol/L. Find the concentration CA after 20 min.
k = 0.5 L/(mol·min)CA0 = 2 mol/Lt = 20 min
CA = mol/L
14.Reaction kineticsHard
The rate constant of a reaction is 0.01 min⁻¹ at 20 °C and 0.0295 min⁻¹ at 30 °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(C3H8) = 24/44.1 = 0.5442 mol
n(O2) = 64/32 = 2 mol
Limiting reactant
0.5442/1 = 0.5442, 2/5 = 0.4 → O2
Mass of product
n(CO2) = 0.4·3 = 1.2 mol
m = 1.2·44.01 = 52.81 g
Answer
m = 52.81 g
2.Stoichiometry
Answer: Y = 78.08 %
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(CaCO3) = 8/100.09 = 0.07993 mol
m = 0.07993·1/1·56.08 = 4.482 g
Percent yield
Y = 3.5/4.482·100 = 78.08%
Answer
Y = 78.08%
3.Molarity & dilution
Answer: Vwater = 450 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 = 0.5·500/5 = 50 mL
Water to add
Vwater = 500 − 50 = 450 mL
Answer
Vwater = 450 mL
4.Molarity & dilution
Answer: c = 0.38 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 = 1.5·100 + 0.1·400 = 190 mmol
Concentration of the mixture
c = 190/500 = 0.38 mol/L
Answer
c = 0.38 mol/L
5.Ideal gas law
Answer: P = 1530 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 430/16.04 = 26.81 mol
T = 70 + 273.15 = 343.1 K
Pressure
P = nRT/V = 26.81·8.314·343.1/0.05 = 1530 kPa
Answer
P = 1530 kPa
6.Ideal gas law
Answer: V2 = 17.26 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Absolute temperatures
T1 = 303.1 K, T2 = 523.1 K
Combined gas law
V2 = V1·(P1/P2)·(T2/T1) = 20·(200/400)·(523.1/303.1) = 17.26 L
Answer
V2 = 17.26 L
7.Calorimetry & enthalpy
Answer: Tf = 33.57 °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
150·0.897·(170 − Tf) = 250·4.18·(Tf − 16)
Solve for the final temperature
Tf = (134.6·170 + 1045·16)/(134.6 + 1045) = 33.57 °C
Answer
Tf = 33.57 °C
8.Calorimetry & enthalpy
Answer: ΔH° = −3119 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] = 0.3111 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.
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
0.08 = x²/(1 − x) → x² + 0.08·x − 0.08 = 0
Solve the quadratic
x = (−0.08 + √(0.08² + 4·0.08))/2 = 0.2457 mol/L
Equilibrium concentration
[Cl₂] = x = 0.2457 mol/L
Answer
[Cl2] = 0.2457 mol/L
11.pH & buffers
Answer: pH = 3.027
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/(0.05 − x) → x = 0.0009397 mol/L
pH
pH = −log(0.0009397) = 3.027
Answer
pH = 3.027
12.pH & buffers
Answer: pH = 3.866
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(0.5/0.1) = 3.866
Answer
pH = 3.866
13.Reaction kinetics
Answer: CA = 0.09524 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 = 0.5 + 0.5·20 = 10.5 L/mol
CA = 1/10.5 = 0.09524 mol/L
Answer
CA = 0.09524 mol/L
14.Reaction kinetics
Answer: Ea = 79.93 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 = 293.1 K, T2 = 303.1 K
Solve Arrhenius for Ea
Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·1.082/(-1.125e-4) = 79929 J/mol