10 g of Na and 32 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 2Al + 3Cl2 → 2AlCl3, 4 g of Al reacts completely (the other reactants are in excess) and 16.8 g of AlCl3 is collected. Find the percent yield Y.
M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 16.8 g
Y = %
3.Molarity & dilutionMedium
You need 100 mL of 0.5 mol/L solution and have a 3 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 3 mol/Lc2 = 0.5 mol/LV2 = 100 mL
V1 = mL
4.Molarity & dilutionHard
200 mL of a 0.25 mol/L solution is mixed with 250 mL of a 0.6 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 250 g block of copper (c = 0.385 J/(g·°C)) at 200 °C is dropped into 200 g of water (c = 4.18 J/(g·°C)) at 25 °C in an insulated cup. Find the final temperature Tf.
0.2 mol/L of H₂ and 0.2 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 16. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.2 mol/LKc = 16
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.01 at the reaction temperature. A container starts with 0.4 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
[PCl₅]₀ = 0.4 mol/LKc = 0.01
[PCl5] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.5 mol/L solution of hydrofluoric acid (Ka = 0.00068).
c = 0.5 mol/LKa = 0.00068
pH =
12.pH & buffersHard
A buffer contains 0.2 mol/L hypochlorous acid and 0.1 mol/L of its sodium salt. Ka = 3 × 10⁻⁸. Find the pH pH of the buffer.
[HA] = 0.2[A⁻] = 0.1Ka = 3 × 10⁻⁸
pH =
13.Reaction kineticsMedium
A reaction with rate law −rA = k·CA² has k = 0.25 L/(mol·min) and starts at 0.5 mol/L. Find the concentration CA after 10 min.
k = 0.25 L/(mol·min)CA0 = 0.5 mol/Lt = 10 min
CA = mol/L
14.Reaction kineticsHard
A reaction has a rate constant of 0.005 min⁻¹ at 25 °C and an activation energy of 60 kJ/mol. Find the rate constant k2 at 55 °C.
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) = 10/22.99 = 0.435 mol
n(Cl2) = 32/70.9 = 0.4513 mol
Limiting reactant
0.435/2 = 0.2175, 0.4513/1 = 0.4513 → Na
Mass of product
n(NaCl) = 0.2175·2 = 0.435 mol
m = 0.435·58.44 = 25.42 g
Answer
m = 25.42 g
2.Stoichiometry
Answer: Y = 84.99 %
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) = 4/26.98 = 0.1483 mol
m = 0.1483·2/2·133.33 = 19.77 g
Percent yield
Y = 16.8/19.77·100 = 84.99%
Answer
Y = 84.99%
3.Molarity & dilution
Answer: V1 = 16.67 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 = 0.5·100/3 = 16.67 mL
Answer
V1 = 16.67 mL
4.Molarity & dilution
Answer: c = 0.4444 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 = 0.25·200 + 0.6·250 = 200 mmol
Concentration of the mixture
c = 200/450 = 0.4444 mol/L
Answer
c = 0.4444 mol/L
5.Ideal gas law
Answer: V = 57.46 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 300/39.95 = 7.509 mol
T = 95 + 273.15 = 368.1 K
Volume
V = nRT/P = 7.509·8.314·368.1/400 = 57.46 L
Answer
V = 57.46 L
6.Ideal gas law
Answer: ρ = 5.971 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 150 + 273.15 = 423.1 K
Density
ρ = 750·28.01/(8.314·423.1) = 5.971 kg/m³
Answer
ρ = 5.971 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 43.07 °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
250·0.385·(200 − Tf) = 200·4.18·(Tf − 25)
Solve for the final temperature
Tf = (96.25·200 + 836·25)/(96.25 + 836) = 43.07 °C
Answer
Tf = 43.07 °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: [H2] = 0.06667 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/(0.4 − x) → x² + 0.01·x − 0.004 = 0
Solve the quadratic
x = (−0.01 + √(0.01² + 4·0.004))/2 = 0.05844 mol/L
Equilibrium concentration
[PCl₅] = 0.4 − 0.05844 = 0.3416 mol/L
Answer
[PCl5] = 0.3416 mol/L
11.pH & buffers
Answer: pH = 1.742
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/(0.5 − x) → x = 0.0181 mol/L
pH
pH = −log(0.0181) = 1.742
Answer
pH = 1.742
12.pH & buffers
Answer: pH = 7.222
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 = 7.523
Henderson–Hasselbalch
pH = 7.523 + log(0.1/0.2) = 7.222
Answer
pH = 7.222
13.Reaction kinetics
Answer: CA = 0.2222 mol/L
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).
Integrated second-order law
1/CA = 1/CA0 + kt = 2 + 0.25·10 = 4.5 L/mol
CA = 1/4.5 = 0.2222 mol/L
Answer
CA = 0.2222 mol/L
14.Reaction kinetics
Answer: k2 = 0.04571 min⁻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).