12 g of Al and 40 g of Cl2 react according to 2Al + 3Cl2 → 2AlCl3. What mass m of AlCl3 can form?
M(Al) = 26.98M(Cl2) = 70.9M(AlCl3) = 133.33 g/mol
m = g
2.StoichiometryHard
In the reaction 2Al + 3Cl2 → 2AlCl3, 16 g of Al reacts completely (the other reactants are in excess) and 64.84 g of AlCl3 is collected. Find the percent yield Y.
M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 64.84 g
Y = %
3.Molarity & dilutionMedium
You need 100 mL of 0.05 mol/L solution and have a 4 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 4 mol/Lc2 = 0.05 mol/LV2 = 100 mL
V1 = mL
4.Molarity & dilutionHard
150 mL of a 0.8 mol/L solution is mixed with 400 mL of a 0.05 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 75 g block of zinc (c = 0.388 J/(g·°C)) at 80 °C is dropped into 400 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 10 g of C2H6 (M = 30.07 g/mol) burns completely: 2C2H6 + 7O₂ → 4CO₂ + 6H₂O(l). Give q as a positive number.
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 = 64. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.5 mol/LKc = 64
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.05 at the reaction temperature. A container starts with 1 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
[PCl₅]₀ = 1 mol/LKc = 0.05
[PCl5] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.01 mol/L solution of lactic acid (Ka = 0.00014).
c = 0.01 mol/LKa = 0.00014
pH =
12.pH & buffersHard
A 1.00 L buffer contains 0.1 mol hydrofluoric acid and 0.2 mol of its conjugate base (Ka = 0.00068). 0.05 mol of NaOH is added. Find the new pH pH.
A first-order reaction in a batch reactor has k = 0.08 min⁻¹. How long does it take to reach 99% conversion? Find t.
k = 0.08 min⁻¹X = 99 %
t = min
14.Reaction kineticsHard
A first-order reaction has a rate constant of 0.01 s⁻¹ at 80 °C and an activation energy of 80 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(Al) = 12/26.98 = 0.4448 mol
n(Cl2) = 40/70.9 = 0.5642 mol
Limiting reactant
0.4448/2 = 0.2224, 0.5642/3 = 0.1881 → Cl2
Mass of product
n(AlCl3) = 0.1881·2 = 0.3761 mol
m = 0.3761·133.33 = 50.15 g
Answer
m = 50.15 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(Al) = 16/26.98 = 0.593 mol
m = 0.593·2/2·133.33 = 79.07 g
Percent yield
Y = 64.84/79.07·100 = 82%
Answer
Y = 82%
3.Molarity & dilution
Answer: V1 = 1.25 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 = 0.05·100/4 = 1.25 mL
Answer
V1 = 1.25 mL
4.Molarity & dilution
Answer: c = 0.2545 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 = 0.8·150 + 0.05·400 = 140 mmol
Concentration of the mixture
c = 140/550 = 0.2545 mol/L
Answer
c = 0.2545 mol/L
5.Ideal gas law
Answer: P = 556.2 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 420/39.95 = 10.51 mol
T = 45 + 273.15 = 318.1 K
Pressure
P = nRT/V = 10.51·8.314·318.1/0.05 = 556.2 kPa
Answer
P = 556.2 kPa
6.Ideal gas law
Answer: ρ = 5.016 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 110 + 273.15 = 383.1 K
Density
ρ = 400·39.95/(8.314·383.1) = 5.016 kg/m³
Answer
ρ = 5.016 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 21.03 °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
75·0.388·(80 − Tf) = 400·4.18·(Tf − 20)
Solve for the final temperature
Tf = (29.1·80 + 1672·20)/(29.1 + 1672) = 21.03 °C
Answer
Tf = 21.03 °C
8.Calorimetry & enthalpy
Answer: q = 518.7 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.
Equilibrium expression
64 = (2x)²/(0.5 − x)²
Take the square root
8 = 2x/(0.5 − x) → x = 0.5·8/(2 + 8) = 0.4 mol/L
Equilibrium concentration
[H₂] = 0.5 − x = 0.1 mol/L
Answer
[H2] = 0.1 mol/L
10.Chemical equilibrium
Answer: [PCl5] = 0.8 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.
Equilibrium expression
0.05 = x²/(1 − x) → x² + 0.05·x − 0.05 = 0
Solve the quadratic
x = (−0.05 + √(0.05² + 4·0.05))/2 = 0.2 mol/L
Equilibrium concentration
[PCl₅] = 1 − 0.2 = 0.8 mol/L
Answer
[PCl5] = 0.8 mol/L
11.pH & buffers
Answer: pH = 2.953
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 = 0.001115 mol/L
pH
pH = −log(0.001115) = 2.953
Answer
pH = 2.953
12.pH & buffers
Answer: pH = 3.866
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⁻) = 0.2 + 0.05 = 0.25 mol
Henderson–Hasselbalch
pH = 3.167 + log(0.25/0.05) = 3.866
Answer
pH = 3.866
13.Reaction kinetics
Answer: t = 57.56 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)/0.08 = 57.56 min
Answer
t = 57.56 min
14.Reaction kinetics
Answer: A = 6.812 × 10⁹ 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).