In the reaction C3H8 + 5O2 → 3CO2 + 4H2O, 10 g of C3H8 reacts completely (the other reactants are in excess) and 12.75 g of H2O is collected. Find the percent yield Y.
M(C3H8) = 44.1 g/molM(H2O) = 18.02 g/molactual: 12.75 g
Y = %
3.Molarity & dilutionMedium
You need 200 mL of 0.5 mol/L solution and have a 2 mol/L stock solution. How much water Vwater do you add to the stock solution?
c1 = 2 mol/Lc2 = 0.5 mol/LV2 = 200 mL
Vwater = mL
4.Molarity & dilutionHard
250 mL of a 0.8 mol/L solution is mixed with 50 mL of a 1.2 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
c1 = 0.8 mol/LV1 = 250 mLc2 = 1.2 mol/LV2 = 50 mL
c = mol/L
5.Ideal gas lawMedium
A tank of 10 L contains 200 g of CH₄ at 5 °C. Treat it as an ideal gas and find the pressure P.
A 100 g block of aluminium (c = 0.897 J/(g·°C)) at 110 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 22 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find the heat q released when 20 g of C2H5OH (M = 46.07 g/mol) burns completely: C2H5OH + 3O₂ → 2CO₂ + 3H₂O(l). Give q as a positive number.
0.4 mol/L of H₂ and 0.4 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 25. Find the equilibrium concentration [HI].
[H₂]₀ = [I₂]₀ = 0.4 mol/LKc = 25
[HI] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 at the reaction temperature. A container starts with 1.5 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 1.5 mol/LKc = 0.08
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.1 mol/L solution of hypochlorous acid (Ka = 3 × 10⁻⁸).
c = 0.1 mol/LKa = 3 × 10⁻⁸
pH =
12.pH & buffersHard
A 1.00 L buffer contains 0.1 mol hydrofluoric acid and 0.15 mol of its conjugate base (Ka = 0.00068). 0.05 mol of NaOH is added. Find the new pH pH.
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(Fe2O3) = 8/159.69 = 0.0501 mol
n(CO) = 48/28.01 = 1.714 mol
Limiting reactant
0.0501/1 = 0.0501, 1.714/3 = 0.5712 → Fe2O3
Mass of product
n(CO2) = 0.0501·3 = 0.1503 mol
m = 0.1503·44.01 = 6.614 g
Answer
m = 6.614 g
2.Stoichiometry
Answer: Y = 78.01 %
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(C3H8) = 10/44.1 = 0.2268 mol
m = 0.2268·4/1·18.02 = 16.34 g
Percent yield
Y = 12.75/16.34·100 = 78.01%
Answer
Y = 78.01%
3.Molarity & dilution
Answer: Vwater = 150 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 = 0.5·200/2 = 50 mL
Water to add
Vwater = 200 − 50 = 150 mL
Answer
Vwater = 150 mL
4.Molarity & dilution
Answer: c = 0.8667 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 = 0.8·250 + 1.2·50 = 260 mmol
Concentration of the mixture
c = 260/300 = 0.8667 mol/L
Answer
c = 0.8667 mol/L
5.Ideal gas law
Answer: P = 2883 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 200/16.04 = 12.47 mol
T = 5 + 273.15 = 278.1 K
Pressure
P = nRT/V = 12.47·8.314·278.1/0.01 = 2883 kPa
Answer
P = 2883 kPa
6.Ideal gas law
Answer: ρ = 7.553 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 160 + 273.15 = 433.1 K
Density
ρ = 850·32/(8.314·433.1) = 7.553 kg/m³
Answer
ρ = 7.553 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 28.96 °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
100·0.897·(110 − Tf) = 250·4.18·(Tf − 22)
Solve for the final temperature
Tf = (89.7·110 + 1045·22)/(89.7 + 1045) = 28.96 °C
Answer
Tf = 28.96 °C
8.Calorimetry & enthalpy
Answer: q = 593.3 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.
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/(1.5 − x) → x² + 0.08·x − 0.12 = 0
Solve the quadratic
x = (−0.08 + √(0.08² + 4·0.12))/2 = 0.3087 mol/L
Equilibrium concentration
[Cl₂] = x = 0.3087 mol/L
Answer
[Cl2] = 0.3087 mol/L
11.pH & buffers
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]).
Equilibrium
HA ⇌ H⁺ + A⁻: x mol/L of acid ionises.
Ka = x²/(0.1 − x) → x = 5.476 × 10⁻⁵ mol/L
pH
pH = −log(5.476 × 10⁻⁵) = 4.262
Answer
pH = 4.262
12.pH & buffers
Answer: pH = 3.77
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) = 0.1 − 0.05 = 0.05 mol
n(A⁻) = 0.15 + 0.05 = 0.2 mol
Henderson–Hasselbalch
pH = 3.167 + log(0.2/0.05) = 3.77
Answer
pH = 3.77
13.Reaction kinetics
Answer: t = 4.621 min
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).
Conversion and concentration
C/C0 = 1 − X = 0.5
Solve for the time
t = −ln(1 − X)/k = −ln(0.5)/0.15 = 4.621 min
Answer
t = 4.621 min
14.Reaction kinetics
Answer: k2 = 0.125 min⁻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).