12 g of Fe2O3 and 40 g of CO react according to Fe2O3 + 3CO → 2Fe + 3CO2. What mass m of Fe can form?
M(Fe2O3) = 159.69M(CO) = 28.01M(Fe) = 55.85 g/mol
m = g
2.StoichiometryHard
In the reaction C3H8 + 5O2 → 3CO2 + 4H2O, 16 g of C3H8 reacts completely (the other reactants are in excess) and 43.59 g of CO2 is collected. Find the percent yield Y.
M(C3H8) = 44.1 g/molM(CO2) = 44.01 g/molactual: 43.59 g
Y = %
3.Molarity & dilutionMedium
You need 500 mL of 0.2 mol/L solution and have a 2.5 mol/L stock solution. How much water Vwater do you add to the stock solution?
c1 = 2.5 mol/Lc2 = 0.2 mol/LV2 = 500 mL
Vwater = mL
4.Molarity & dilutionHard
50 mL of a 1.5 mol/L solution is mixed with 200 mL of a 0.75 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 250 g block of lead (c = 0.129 J/(g·°C)) at 130 °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 = 54. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.8 mol/LKc = 54
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.2 at the reaction temperature. A container starts with 0.5 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 0.5 mol/LKc = 0.2
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.2 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).
c = 0.2 mol/LKa = 6.3 × 10⁻⁵
pH =
12.pH & buffersHard
A buffer contains 0.2 mol/L formic acid and 0.3 mol/L of its sodium salt. Ka = 0.00018. Find the pH pH of the buffer.
[HA] = 0.2[A⁻] = 0.3Ka = 0.00018
pH =
13.Reaction kineticsMedium
A first-order reaction in a batch reactor has k = 0.3 min⁻¹. How long does it take to reach 95% conversion? Find t.
k = 0.3 min⁻¹X = 95 %
t = min
14.Reaction kineticsHard
A first-order reaction has k = 0.005 min⁻¹ at 20 °C and an activation energy of 60 kJ/mol. The batch reactor is now run at 50 °C for 45 min. Find the conversion X reached.
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) = 12/159.69 = 0.07515 mol
n(CO) = 40/28.01 = 1.428 mol
Limiting reactant
0.07515/1 = 0.07515, 1.428/3 = 0.476 → Fe2O3
Mass of product
n(Fe) = 0.07515·2 = 0.1503 mol
m = 0.1503·55.85 = 8.394 g
Answer
m = 8.394 g
2.Stoichiometry
Answer: Y = 91 %
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) = 16/44.1 = 0.3628 mol
m = 0.3628·3/1·44.01 = 47.9 g
Percent yield
Y = 43.59/47.9·100 = 91%
Answer
Y = 91%
3.Molarity & dilution
Answer: Vwater = 460 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 = 0.2·500/2.5 = 40 mL
Water to add
Vwater = 500 − 40 = 460 mL
Answer
Vwater = 460 mL
4.Molarity & dilution
Answer: c = 0.9 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 = 1.5·50 + 0.75·200 = 225 mmol
Concentration of the mixture
c = 225/250 = 0.9 mol/L
Answer
c = 0.9 mol/L
5.Ideal gas law
Answer: P = 4989 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 210/4.003 = 52.46 mol
T = 70 + 273.15 = 343.1 K
Pressure
P = nRT/V = 52.46·8.314·343.1/0.03 = 4989 kPa
Answer
P = 4989 kPa
6.Ideal gas law
Answer: ρ = 6.406 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 140 + 273.15 = 413.1 K
Density
ρ = 500·44.01/(8.314·413.1) = 6.406 kg/m³
Answer
ρ = 6.406 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 19.41 °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
250·0.129·(130 − Tf) = 250·4.18·(Tf − 16)
Solve for the final temperature
Tf = (32.25·130 + 1045·16)/(32.25 + 1045) = 19.41 °C
Answer
Tf = 19.41 °C
8.Calorimetry & enthalpy
Answer: ΔH° = −2802 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] = 0.1712 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
0.2 = x²/(0.5 − x) → x² + 0.2·x − 0.1 = 0
Solve the quadratic
x = (−0.2 + √(0.2² + 4·0.1))/2 = 0.2317 mol/L
Equilibrium concentration
[Cl₂] = x = 0.2317 mol/L
Answer
[Cl2] = 0.2317 mol/L
11.pH & buffers
Answer: pH = 2.454
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/(0.2 − x) → x = 0.003518 mol/L
pH
pH = −log(0.003518) = 2.454
Answer
pH = 2.454
12.pH & buffers
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]).
pKa
pKa = 3.745
Henderson–Hasselbalch
pH = 3.745 + log(0.3/0.2) = 3.921
Answer
pH = 3.921
13.Reaction kinetics
Answer: t = 9.986 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 = 0.05
Solve for the time
t = −ln(1 − X)/k = −ln(0.05)/0.3 = 9.986 min
Answer
t = 9.986 min
14.Reaction kinetics
Answer: X = 89.05 %
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).