12 g of Al and 10 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 C3H8 + 5O2 → 3CO2 + 4H2O, 30 g of C3H8 reacts completely (the other reactants are in excess) and 44.62 g of H2O is collected. Find the percent yield Y.
M(C3H8) = 44.1 g/molM(H2O) = 18.02 g/molactual: 44.62 g
Y = %
3.Molarity & dilutionMedium
You need 100 mL of 0.3 mol/L solution and have a 0.5 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 0.5 mol/Lc2 = 0.3 mol/LV2 = 100 mL
V1 = mL
4.Molarity & dilutionHard
100 mL of a 1 mol/L solution is mixed with 500 mL of a 2.5 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
c1 = 1 mol/LV1 = 100 mLc2 = 2.5 mol/LV2 = 500 mL
c = mol/L
5.Ideal gas lawMedium
A tank of 50 L contains 420 g of CH₄ at 40 °C. Treat it as an ideal gas and find the pressure P.
A 75 g block of copper (c = 0.385 J/(g·°C)) at 100 °C is dropped into 200 g of water (c = 4.18 J/(g·°C)) at 18 °C in an insulated cup. Find the final temperature Tf.
0.25 mol/L of H₂ and 0.25 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 = 0.25 mol/LKc = 54
[HI] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 at the reaction temperature. A container starts with 1 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
[PCl₅]₀ = 1 mol/LKc = 0.08
[PCl5] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.2 mol/L solution of hypochlorous acid (Ka = 3 × 10⁻⁸).
c = 0.2 mol/LKa = 3 × 10⁻⁸
pH =
12.pH & buffersHard
A buffer contains 0.2 mol/L acetic acid and 0.2 mol/L of its sodium salt. Ka = 1.8 × 10⁻⁵. Find the pH pH of the buffer.
[HA] = 0.2[A⁻] = 0.2Ka = 1.8 × 10⁻⁵
pH =
13.Reaction kineticsMedium
A second-order reaction (−rA = k·CA²) in a batch reactor has k = 0.2 L/(mol·min) and starts at 1 mol/L. How long does it take to reach 80% conversion? Find t.
k = 0.2 L/(mol·min)CA0 = 1 mol/LX = 80 %
t = min
14.Reaction kineticsHard
A first-order reaction has a rate constant of 0.1 s⁻¹ at 30 °C and an activation energy of 70 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) = 10/70.9 = 0.141 mol
Limiting reactant
0.4448/2 = 0.2224, 0.141/3 = 0.04701 → Cl2
Mass of product
n(AlCl3) = 0.04701·2 = 0.09403 mol
m = 0.09403·133.33 = 12.54 g
Answer
m = 12.54 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) = 30/44.1 = 0.6803 mol
m = 0.6803·4/1·18.02 = 49.03 g
Percent yield
Y = 44.62/49.03·100 = 91%
Answer
Y = 91%
3.Molarity & dilution
Answer: V1 = 60 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 = 0.3·100/0.5 = 60 mL
Answer
V1 = 60 mL
4.Molarity & dilution
Answer: c = 2.25 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 + 2.5·500 = 1350 mmol
Concentration of the mixture
c = 1350/600 = 2.25 mol/L
Answer
c = 2.25 mol/L
5.Ideal gas law
Answer: P = 1363 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 420/16.04 = 26.18 mol
T = 40 + 273.15 = 313.1 K
Pressure
P = nRT/V = 26.18·8.314·313.1/0.05 = 1363 kPa
Answer
P = 1363 kPa
6.Ideal gas law
Answer: ρ = 9.297 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 40 + 273.15 = 313.1 K
Density
ρ = 550·44.01/(8.314·313.1) = 9.297 kg/m³
Answer
ρ = 9.297 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 20.74 °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·(100 − Tf) = 200·4.18·(Tf − 18)
Solve for the final temperature
Tf = (28.88·100 + 836·18)/(28.88 + 836) = 20.74 °C
Answer
Tf = 20.74 °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: [HI] = 0.393 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
0.08 = x²/(1 − x) → x² + 0.08·x − 0.08 = 0
Solve the quadratic
x = (−0.08 + √(0.08² + 4·0.08))/2 = 0.2457 mol/L
Equilibrium concentration
[PCl₅] = 1 − 0.2457 = 0.7543 mol/L
Answer
[PCl5] = 0.7543 mol/L
11.pH & buffers
Answer: pH = 4.111
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 = 7.744 × 10⁻⁵ mol/L
pH
pH = −log(7.744 × 10⁻⁵) = 4.111
Answer
pH = 4.111
12.pH & buffers
Answer: pH = 4.745
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 = 4.745
Henderson–Hasselbalch
pH = 4.745 + log(0.2/0.2) = 4.745
Answer
pH = 4.745
13.Reaction kinetics
Answer: t = 20 min
Integrated rate laws: zero order: C = C₀ − kt (straight line, t½ = C₀/(2k)); first order: ln(C₀/C) = kt (t½ = ln 2/k); second order: 1/C − 1/C₀ = kt (t½ = 1/(kC₀)). Conversion X means C = C₀(1 − X).
Concentration at the end
CA = 1·(1 − 0.8) = 0.2 mol/L
Integrated second-order law
t = (1/CA − 1/CA0)/k = (5 − 1)/0.2 = 20 min
Answer
t = 20 min
14.Reaction kinetics
Answer: A = 1.153 × 10¹¹ 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).