8 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 2H2 + O2 → 2H2O, 16 g of H2 reacts completely (the other reactants are in excess) and 134.4 g of H2O is collected. Find the percent yield Y.
M(H2) = 2.016 g/molM(H2O) = 18.02 g/molactual: 134.4 g
Y = %
3.Molarity & dilutionMedium
You need 200 mL of 0.3 mol/L solution and have a 2 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 2 mol/Lc2 = 0.3 mol/LV2 = 200 mL
V1 = mL
4.Molarity & dilutionHard
100 mL of a 0.5 mol/L solution is mixed with 250 mL of a 2.5 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 100 g block of zinc (c = 0.388 J/(g·°C)) at 90 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 21 °C in an insulated cup. Find the final temperature Tf.
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 = 36. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.5 mol/LKc = 36
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 at the reaction temperature. A container starts with 0.4 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 0.4 mol/LKc = 0.08
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.05 mol/L solution of hydrofluoric acid (Ka = 0.00068).
c = 0.05 mol/LKa = 0.00068
pH =
12.pH & buffersHard
A 1.00 L buffer contains 0.3 mol acetic acid and 0.2 mol of its conjugate base (Ka = 1.8 × 10⁻⁵). 0.02 mol of NaOH is added. Find the new pH pH.
A reaction with rate law −rA = k·CA² has k = 0.5 L/(mol·min) and starts at 1 mol/L. Find the concentration CA after 5 min.
k = 0.5 L/(mol·min)CA0 = 1 mol/Lt = 5 min
CA = mol/L
14.Reaction kineticsHard
A first-order reaction has a rate constant of 0.05 s⁻¹ at 25 °C and an activation energy of 100 kJ/mol. Find the pre-exponential factor A in the Arrhenius equation.
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) = 8/44.1 = 0.1814 mol
n(O2) = 64/32 = 2 mol
Limiting reactant
0.1814/1 = 0.1814, 2/5 = 0.4 → C3H8
Mass of product
n(CO2) = 0.1814·3 = 0.5442 mol
m = 0.5442·44.01 = 23.95 g
Answer
m = 23.95 g
2.Stoichiometry
Answer: Y = 94 %
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(H2) = 16/2.016 = 7.937 mol
m = 7.937·2/2·18.02 = 143 g
Percent yield
Y = 134.4/143·100 = 94%
Answer
Y = 94%
3.Molarity & dilution
Answer: V1 = 30 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 = 0.3·200/2 = 30 mL
Answer
V1 = 30 mL
4.Molarity & dilution
Answer: c = 1.929 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 = 0.5·100 + 2.5·250 = 675 mmol
Concentration of the mixture
c = 675/350 = 1.929 mol/L
Answer
c = 1.929 mol/L
5.Ideal gas law
Answer: P = 50.38 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 30/32 = 0.9375 mol
T = 50 + 273.15 = 323.1 K
Pressure
P = nRT/V = 0.9375·8.314·323.1/0.05 = 50.38 kPa
Answer
P = 50.38 kPa
6.Ideal gas law
Answer: ρ = 0.7277 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 190 + 273.15 = 463.1 K
Density
ρ = 700·4.003/(8.314·463.1) = 0.7277 kg/m³
Answer
ρ = 0.7277 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 23.47 °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
100·0.388·(90 − Tf) = 250·4.18·(Tf − 21)
Solve for the final temperature
Tf = (38.8·90 + 1045·21)/(38.8 + 1045) = 23.47 °C
Answer
Tf = 23.47 °C
8.Calorimetry & enthalpy
Answer: ΔH° = −1453 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: [H2] = 0.125 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/(0.4 − x) → x² + 0.08·x − 0.032 = 0
Solve the quadratic
x = (−0.08 + √(0.08² + 4·0.032))/2 = 0.1433 mol/L
Equilibrium concentration
[Cl₂] = x = 0.1433 mol/L
Answer
[Cl2] = 0.1433 mol/L
11.pH & buffers
Answer: pH = 2.26
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 = 0.005501 mol/L
pH
pH = −log(0.005501) = 2.26
Answer
pH = 2.26
12.pH & buffers
Answer: pH = 4.64
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]).
Neutralisation
n(HA) = 0.3 − 0.02 = 0.28 mol
n(A⁻) = 0.2 + 0.02 = 0.22 mol
Henderson–Hasselbalch
pH = 4.745 + log(0.22/0.28) = 4.64
Answer
pH = 4.64
13.Reaction kinetics
Answer: CA = 0.2857 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 = 1 + 0.5·5 = 3.5 L/mol
CA = 1/3.5 = 0.2857 mol/L
Answer
CA = 0.2857 mol/L
14.Reaction kinetics
Answer: A = 1.656 × 10¹⁶ s⁻¹
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).