24 g of Na and 16 g of Cl2 react according to 2Na + Cl2 → 2NaCl. What mass m of NaCl can form?
M(Na) = 22.99M(Cl2) = 70.9M(NaCl) = 58.44 g/mol
m = g
2.StoichiometryHard
In the reaction 2Na + Cl2 → 2NaCl, 30 g of Na reacts completely (the other reactants are in excess) and 62.53 g of NaCl is collected. Find the percent yield Y.
M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 62.53 g
Y = %
3.Molarity & dilutionMedium
You need 200 mL of 0.5 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 = 0.5 mol/LV2 = 200 mL
Vwater = mL
4.Molarity & dilutionHard
100 mL of a 1 mol/L solution is mixed with 250 mL of a 0.75 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
c1 = 1 mol/LV1 = 100 mLc2 = 0.75 mol/LV2 = 250 mL
c = mol/L
5.Ideal gas lawMedium
What volume V does 150 g of Ar occupy at 500 kPa and 95 °C? Treat it as an ideal gas.
A 75 g block of copper (c = 0.385 J/(g·°C)) at 90 °C is dropped into 400 g of water (c = 4.18 J/(g·°C)) at 17 °C in an insulated cup. Find the final temperature Tf.
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 = 54. Find the equilibrium concentration [HI].
[H₂]₀ = [I₂]₀ = 0.4 mol/LKc = 54
[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 0.01 mol/L solution of hypochlorous acid (Ka = 3 × 10⁻⁸).
c = 0.01 mol/LKa = 3 × 10⁻⁸
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 first-order reaction in a batch reactor has k = 0.3 min⁻¹. How long does it take to reach 60% conversion? Find t.
k = 0.3 min⁻¹X = 60 %
t = min
14.Reaction kineticsHard
A first-order reaction has k = 0.02 min⁻¹ at 25 °C and an activation energy of 70 kJ/mol. The batch reactor is now run at 55 °C for 20 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(Na) = 24/22.99 = 1.044 mol
n(Cl2) = 16/70.9 = 0.2257 mol
Limiting reactant
1.044/2 = 0.522, 0.2257/1 = 0.2257 → Cl2
Mass of product
n(NaCl) = 0.2257·2 = 0.4513 mol
m = 0.4513·58.44 = 26.38 g
Answer
m = 26.38 g
2.Stoichiometry
Answer: Y = 82 %
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(Na) = 30/22.99 = 1.305 mol
m = 1.305·2/2·58.44 = 76.26 g
Percent yield
Y = 62.53/76.26·100 = 82%
Answer
Y = 82%
3.Molarity & dilution
Answer: Vwater = 160 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 = 0.5·200/2.5 = 40 mL
Water to add
Vwater = 200 − 40 = 160 mL
Answer
Vwater = 160 mL
4.Molarity & dilution
Answer: c = 0.8214 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 = 1·100 + 0.75·250 = 287.5 mmol
Concentration of the mixture
c = 287.5/350 = 0.8214 mol/L
Answer
c = 0.8214 mol/L
5.Ideal gas law
Answer: V = 22.98 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 150/39.95 = 3.755 mol
T = 95 + 273.15 = 368.1 K
Volume
V = nRT/P = 3.755·8.314·368.1/500 = 22.98 L
Answer
V = 22.98 L
6.Ideal gas law
Answer: ρ = 3.291 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 20 + 273.15 = 293.1 K
Density
ρ = 500·16.04/(8.314·293.1) = 3.291 kg/m³
Answer
ρ = 3.291 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 18.24 °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·(90 − Tf) = 400·4.18·(Tf − 17)
Solve for the final temperature
Tf = (28.88·90 + 1672·17)/(28.88 + 1672) = 18.24 °C
Answer
Tf = 18.24 °C
8.Calorimetry & enthalpy
Answer: ΔH° = −890.3 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.
Negative: the reaction gives off heat (exothermic).
9.Chemical equilibrium
Answer: [HI] = 0.6288 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
0.01 = 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 = 4.762
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/(0.01 − x) → x = 1.731 × 10⁻⁵ mol/L
pH
pH = −log(1.731 × 10⁻⁵) = 4.762
Answer
pH = 4.762
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: t = 3.054 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 = 0.4
Solve for the time
t = −ln(1 − X)/k = −ln(0.4)/0.3 = 3.054 min
Answer
t = 3.054 min
14.Reaction kinetics
Answer: X = 99.49 %
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).