16 g of Na and 40 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 N2 + 3H2 → 2NH3, 12 g of N2 reacts completely (the other reactants are in excess) and 8.02 g of NH3 is collected. Find the percent yield Y.
M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 8.02 g
Y = %
3.Molarity & dilutionMedium
You need 200 mL of 0.25 mol/L solution and have a 2 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 2 mol/Lc2 = 0.25 mol/LV2 = 200 mL
V1 = mL
4.Molarity & dilutionHard
50 mL of a 0.8 mol/L solution is mixed with 400 mL of a 0.6 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
c1 = 0.8 mol/LV1 = 50 mLc2 = 0.6 mol/LV2 = 400 mL
c = mol/L
5.Ideal gas lawMedium
A tank of 30 L contains 470 g of Ar at 85 °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 160 °C is dropped into 200 g of water (c = 4.18 J/(g·°C)) at 20 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find the heat q released when 250 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 0.5 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 0.5 mol/LKc = 0.08
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.25 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).
c = 0.25 mol/LKa = 6.3 × 10⁻⁵
pH =
12.pH & buffersHard
A buffer contains 0.2 mol/L lactic acid and 0.05 mol/L of its sodium salt. Ka = 0.00014. Find the pH pH of the buffer.
[HA] = 0.2[A⁻] = 0.05Ka = 0.00014
pH =
13.Reaction kineticsMedium
A first-order reaction in a batch reactor has k = 0.01 min⁻¹. How long does it take to reach 99% conversion? Find t.
k = 0.01 min⁻¹X = 99 %
t = min
14.Reaction kineticsHard
The rate constant of a reaction is 0.01 min⁻¹ at 20 °C and 0.0271 min⁻¹ at 35 °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(Na) = 16/22.99 = 0.696 mol
n(Cl2) = 40/70.9 = 0.5642 mol
Limiting reactant
0.696/2 = 0.348, 0.5642/1 = 0.5642 → Na
Mass of product
n(NaCl) = 0.348·2 = 0.696 mol
m = 0.696·58.44 = 40.67 g
Answer
m = 40.67 g
2.Stoichiometry
Answer: Y = 54.98 %
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(N2) = 12/28.02 = 0.4283 mol
m = 0.4283·2/1·17.03 = 14.59 g
Percent yield
Y = 8.02/14.59·100 = 54.98%
Answer
Y = 54.98%
3.Molarity & dilution
Answer: V1 = 25 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 = 0.25·200/2 = 25 mL
Answer
V1 = 25 mL
4.Molarity & dilution
Answer: c = 0.6222 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 = 0.8·50 + 0.6·400 = 280 mmol
Concentration of the mixture
c = 280/450 = 0.6222 mol/L
Answer
c = 0.6222 mol/L
5.Ideal gas law
Answer: P = 1168 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 470/39.95 = 11.76 mol
T = 85 + 273.15 = 358.1 K
Pressure
P = nRT/V = 11.76·8.314·358.1/0.03 = 1168 kPa
Answer
P = 1168 kPa
6.Ideal gas law
Answer: V2 = 0.9805 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Absolute temperatures
T1 = 283.1 K, T2 = 333.1 K
Combined gas law
V2 = V1·(P1/P2)·(T2/T1) = 2·(250/600)·(333.1/283.1) = 0.9805 L
Answer
V2 = 0.9805 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
100·0.897·(160 − Tf) = 200·4.18·(Tf − 20)
Solve for the final temperature
Tf = (89.7·160 + 836·20)/(89.7 + 836) = 33.57 °C
Answer
Tf = 33.57 °C
8.Calorimetry & enthalpy
Answer: q = 7416 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/(0.5 − x) → x² + 0.08·x − 0.04 = 0
Solve the quadratic
x = (−0.08 + √(0.08² + 4·0.04))/2 = 0.164 mol/L
Equilibrium concentration
[Cl₂] = x = 0.164 mol/L
Answer
[Cl2] = 0.164 mol/L
11.pH & buffers
Answer: pH = 2.405
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/(0.25 − x) → x = 0.003937 mol/L
pH
pH = −log(0.003937) = 2.405
Answer
pH = 2.405
12.pH & buffers
Answer: pH = 3.252
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 = 3.854
Henderson–Hasselbalch
pH = 3.854 + log(0.05/0.2) = 3.252
Answer
pH = 3.252
13.Reaction kinetics
Answer: t = 460.5 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 = 0.01
Solve for the time
t = −ln(1 − X)/k = −ln(0.01)/0.01 = 460.5 min
Answer
t = 460.5 min
14.Reaction kinetics
Answer: Ea = 49.92 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 = 308.1 K
Solve Arrhenius for Ea
Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·0.9969/(-1.661e-4) = 49916 J/mol