4 g of C3H8 and 64 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of H2O can form?
M(C3H8) = 44.1M(O2) = 32M(H2O) = 18.02 g/mol
m = g
2.StoichiometryHard
In the reaction C6H12O6 → 2C2H5OH + 2CO2, 8 g of C6H12O6 reacts completely (the other reactants are in excess) and 3.48 g of C2H5OH is collected. Find the percent yield Y.
M(C6H12O6) = 180.16 g/molM(C2H5OH) = 46.07 g/molactual: 3.48 g
Y = %
3.Molarity & dilutionMedium
You need 250 mL of 0.4 mol/L solution and have a 3 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 3 mol/Lc2 = 0.4 mol/LV2 = 250 mL
V1 = mL
4.Molarity & dilutionHard
150 mL of a 2 mol/L solution is mixed with 100 mL of a 0.75 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
c1 = 2 mol/LV1 = 150 mLc2 = 0.75 mol/LV2 = 100 mL
c = mol/L
5.Ideal gas lawMedium
What volume V does 160 g of CO₂ occupy at 450 kPa and 20 °C? Treat it as an ideal gas.
A 200 g block of iron (c = 0.449 J/(g·°C)) at 200 °C is dropped into 400 g of water (c = 4.18 J/(g·°C)) at 18 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find the heat q released when 100 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 = 36. Find the equilibrium concentration [HI].
[H₂]₀ = [I₂]₀ = 0.5 mol/LKc = 36
[HI] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.5 at the reaction temperature. A container starts with 1.5 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
[PCl₅]₀ = 1.5 mol/LKc = 0.5
[PCl5] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.5 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).
c = 0.5 mol/LKa = 6.3 × 10⁻⁵
pH =
12.pH & buffersHard
A 1.00 L buffer contains 0.3 mol hydrofluoric acid and 0.1 mol of its conjugate base (Ka = 0.00068). 0.05 mol of NaOH is added. Find the new pH pH.
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(C3H8) = 4/44.1 = 0.0907 mol
n(O2) = 64/32 = 2 mol
Limiting reactant
0.0907/1 = 0.0907, 2/5 = 0.4 → C3H8
Mass of product
n(H2O) = 0.0907·4 = 0.3628 mol
m = 0.3628·18.02 = 6.538 g
Answer
m = 6.538 g
2.Stoichiometry
Answer: Y = 85.05 %
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(C6H12O6) = 8/180.16 = 0.0444 mol
m = 0.0444·2/1·46.07 = 4.091 g
Percent yield
Y = 3.48/4.091·100 = 85.05%
Answer
Y = 85.05%
3.Molarity & dilution
Answer: V1 = 33.33 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 = 0.4·250/3 = 33.33 mL
Answer
V1 = 33.33 mL
4.Molarity & dilution
Answer: c = 1.5 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 = 2·150 + 0.75·100 = 375 mmol
Concentration of the mixture
c = 375/250 = 1.5 mol/L
Answer
c = 1.5 mol/L
5.Ideal gas law
Answer: V = 19.69 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 160/44.01 = 3.636 mol
T = 20 + 273.15 = 293.1 K
Volume
V = nRT/P = 3.636·8.314·293.1/450 = 19.69 L
Answer
V = 19.69 L
6.Ideal gas law
Answer: ρ = 1.059 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 0 + 273.15 = 273.1 K
Density
ρ = 150·16.04/(8.314·273.1) = 1.059 kg/m³
Answer
ρ = 1.059 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 27.28 °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
200·0.449·(200 − Tf) = 400·4.18·(Tf − 18)
Solve for the final temperature
Tf = (89.8·200 + 1672·18)/(89.8 + 1672) = 27.28 °C
Answer
Tf = 27.28 °C
8.Calorimetry & enthalpy
Answer: q = 5187 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
0.5 = x²/(1.5 − x) → x² + 0.5·x − 0.75 = 0
Solve the quadratic
x = (−0.5 + √(0.5² + 4·0.75))/2 = 0.6514 mol/L
Equilibrium concentration
[PCl₅] = 1.5 − 0.6514 = 0.8486 mol/L
Answer
[PCl5] = 0.8486 mol/L
11.pH & buffers
Answer: pH = 2.253
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 = 0.005581 mol/L
pH
pH = −log(0.005581) = 2.253
Answer
pH = 2.253
12.pH & buffers
Answer: pH = 2.946
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) = 0.3 − 0.05 = 0.25 mol
n(A⁻) = 0.1 + 0.05 = 0.15 mol
Henderson–Hasselbalch
pH = 3.167 + log(0.15/0.25) = 2.946
Answer
pH = 2.946
13.Reaction kinetics
Answer: t = 17.33 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 = 0.25
Solve for the time
t = −ln(1 − X)/k = −ln(0.25)/0.08 = 17.33 min
Answer
t = 17.33 min
14.Reaction kinetics
Answer: Ea = 100.1 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 = 313.1 K, T2 = 333.1 K
Solve Arrhenius for Ea
Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·2.308/(-1.917e-4) = 100075 J/mol