10 g of CH4 and 32 g of O2 react according to CH4 + 2O2 → CO2 + 2H2O. What mass m of CO2 can form?
M(CH4) = 16.04M(O2) = 32M(CO2) = 44.01 g/mol
m = g
2.StoichiometryHard
In the reaction 2Al + 3Cl2 → 2AlCl3, 12 g of Al reacts completely (the other reactants are in excess) and 46.26 g of AlCl3 is collected. Find the percent yield Y.
M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 46.26 g
Y = %
3.Molarity & dilutionMedium
You need 1000 mL of 0.1 mol/L solution and have a 1.5 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 1.5 mol/Lc2 = 0.1 mol/LV2 = 1000 mL
V1 = mL
4.Molarity & dilutionHard
50 mL of a 0.8 mol/L solution is mixed with 200 mL of a 1.2 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
c1 = 0.8 mol/LV1 = 50 mLc2 = 1.2 mol/LV2 = 200 mL
c = mol/L
5.Ideal gas lawMedium
A tank of 20 L contains 330 g of O₂ at 100 °C. Treat it as an ideal gas and find the pressure P.
A 100 g block of copper (c = 0.385 J/(g·°C)) at 190 °C is dropped into 150 g of water (c = 4.18 J/(g·°C)) at 23 °C in an insulated cup. Find the final temperature Tf.
0.1 mol/L of H₂ and 0.1 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 = 0.1 mol/LKc = 64
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.2 at the reaction temperature. A container starts with 0.2 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 0.2 mol/LKc = 0.2
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.01 mol/L solution of hydrofluoric acid (Ka = 0.00068).
c = 0.01 mol/LKa = 0.00068
pH =
12.pH & buffersHard
A buffer contains 0.4 mol/L lactic acid and 0.8 mol/L of its sodium salt. Ka = 0.00014. Find the pH pH of the buffer.
[HA] = 0.4[A⁻] = 0.8Ka = 0.00014
pH =
13.Reaction kineticsMedium
A reaction with rate law −rA = k·CA² has k = 0.05 L/(mol·min) and starts at 1 mol/L. Find the concentration CA after 30 min.
k = 0.05 L/(mol·min)CA0 = 1 mol/Lt = 30 min
CA = mol/L
14.Reaction kineticsHard
A first-order reaction has a rate constant of 0.05 s⁻¹ at 60 °C and an activation energy of 100 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(CH4) = 10/16.04 = 0.6234 mol
n(O2) = 32/32 = 1 mol
Limiting reactant
0.6234/1 = 0.6234, 1/2 = 0.5 → O2
Mass of product
n(CO2) = 0.5·1 = 0.5 mol
m = 0.5·44.01 = 22 g
Answer
m = 22 g
2.Stoichiometry
Answer: Y = 78.01 %
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) = 12/26.98 = 0.4448 mol
m = 0.4448·2/2·133.33 = 59.3 g
Percent yield
Y = 46.26/59.3·100 = 78.01%
Answer
Y = 78.01%
3.Molarity & dilution
Answer: V1 = 66.67 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 = 0.1·1000/1.5 = 66.67 mL
Answer
V1 = 66.67 mL
4.Molarity & dilution
Answer: c = 1.12 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 + 1.2·200 = 280 mmol
Concentration of the mixture
c = 280/250 = 1.12 mol/L
Answer
c = 1.12 mol/L
5.Ideal gas law
Answer: P = 1600 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 330/32 = 10.31 mol
T = 100 + 273.15 = 373.1 K
Pressure
P = nRT/V = 10.31·8.314·373.1/0.02 = 1600 kPa
Answer
P = 1600 kPa
6.Ideal gas law
Answer: V2 = 10.6 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Absolute temperatures
T1 = 283.1 K, T2 = 343.1 K
Combined gas law
V2 = V1·(P1/P2)·(T2/T1) = 10·(350/400)·(343.1/283.1) = 10.6 L
Answer
V2 = 10.6 L
7.Calorimetry & enthalpy
Answer: Tf = 32.66 °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
100·0.385·(190 − Tf) = 150·4.18·(Tf − 23)
Solve for the final temperature
Tf = (38.5·190 + 627·23)/(38.5 + 627) = 32.66 °C
Answer
Tf = 32.66 °C
8.Calorimetry & enthalpy
Answer: ΔH° = −5755 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] = 0.02 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
64 = (2x)²/(0.1 − x)²
Take the square root
8 = 2x/(0.1 − x) → x = 0.1·8/(2 + 8) = 0.08 mol/L
Equilibrium concentration
[H₂] = 0.1 − x = 0.02 mol/L
Answer
[H2] = 0.02 mol/L
10.Chemical equilibrium
Answer: [Cl2] = 0.1236 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
0.2 = x²/(0.2 − x) → x² + 0.2·x − 0.04 = 0
Solve the quadratic
x = (−0.2 + √(0.2² + 4·0.04))/2 = 0.1236 mol/L
Equilibrium concentration
[Cl₂] = x = 0.1236 mol/L
Answer
[Cl2] = 0.1236 mol/L
11.pH & buffers
Answer: pH = 2.64
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 = 0.00229 mol/L
pH
pH = −log(0.00229) = 2.64
Answer
pH = 2.64
12.pH & buffers
Answer: pH = 4.155
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(0.8/0.4) = 4.155
Answer
pH = 4.155
13.Reaction kinetics
Answer: CA = 0.4 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 = 1 + 0.05·30 = 2.5 L/mol
CA = 1/2.5 = 0.4 mol/L
Answer
CA = 0.4 mol/L
14.Reaction kinetics
Answer: A = 2.391 × 10¹⁴ 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).