8 g of Mg and 16 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?
M(Mg) = 24.31M(O2) = 32M(MgO) = 40.3 g/mol
m = g
2.StoichiometryHard
In the reaction CH4 + 2O2 → CO2 + 2H2O, 6 g of CH4 reacts completely (the other reactants are in excess) and 12.35 g of CO2 is collected. Find the percent yield Y.
M(CH4) = 16.04 g/molM(CO2) = 44.01 g/molactual: 12.35 g
Y = %
3.Molarity & dilutionMedium
You need 500 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 = 500 mL
V1 = mL
4.Molarity & dilutionHard
100 mL of a 0.4 mol/L solution is mixed with 400 mL of a 2.5 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 200 g block of copper (c = 0.385 J/(g·°C)) at 110 °C is dropped into 150 g of water (c = 4.18 J/(g·°C)) at 24 °C in an insulated cup. Find the final temperature Tf.
1 mol/L of H₂ and 1 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 50. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 1 mol/LKc = 50
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.1 at the reaction temperature. A container starts with 0.1 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 0.1 mol/LKc = 0.1
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.05 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).
c = 0.05 mol/LKa = 6.3 × 10⁻⁵
pH =
12.pH & buffersHard
A buffer contains 0.5 mol/L acetic acid and 0.1 mol/L of its sodium salt. Ka = 1.8 × 10⁻⁵. Find the pH pH of the buffer.
[HA] = 0.5[A⁻] = 0.1Ka = 1.8 × 10⁻⁵
pH =
13.Reaction kineticsMedium
A first-order reaction in a batch reactor has k = 0.3 min⁻¹. How long does it take to reach 60% conversion? Find t.
k = 0.3 min⁻¹X = 60 %
t = min
14.Reaction kineticsHard
A reaction has a rate constant of 0.05 min⁻¹ at 30 °C and an activation energy of 55 kJ/mol. Find the rate constant k2 at 60 °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(Mg) = 8/24.31 = 0.3291 mol
n(O2) = 16/32 = 0.5 mol
Limiting reactant
0.3291/2 = 0.1645, 0.5/1 = 0.5 → Mg
Mass of product
n(MgO) = 0.1645·2 = 0.3291 mol
m = 0.3291·40.3 = 13.26 g
Answer
m = 13.26 g
2.Stoichiometry
Answer: Y = 75.02 %
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(CH4) = 6/16.04 = 0.3741 mol
m = 0.3741·1/1·44.01 = 16.46 g
Percent yield
Y = 12.35/16.46·100 = 75.02%
Answer
Y = 75.02%
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 = 0.1·500/1.5 = 33.33 mL
Answer
V1 = 33.33 mL
4.Molarity & dilution
Answer: c = 2.08 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 = 0.4·100 + 2.5·400 = 1040 mmol
Concentration of the mixture
c = 1040/500 = 2.08 mol/L
Answer
c = 2.08 mol/L
5.Ideal gas law
Answer: V = 320.2 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 400/32 = 12.5 mol
T = 35 + 273.15 = 308.1 K
Volume
V = nRT/P = 12.5·8.314·308.1/100 = 320.2 L
Answer
V = 320.2 L
6.Ideal gas law
Answer: ρ = 6.521 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 140 + 273.15 = 413.1 K
Density
ρ = 700·32/(8.314·413.1) = 6.521 kg/m³
Answer
ρ = 6.521 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 33.41 °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
200·0.385·(110 − Tf) = 150·4.18·(Tf − 24)
Solve for the final temperature
Tf = (77·110 + 627·24)/(77 + 627) = 33.41 °C
Answer
Tf = 33.41 °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: [H2] = 0.2205 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
0.1 = x²/(0.1 − x) → x² + 0.1·x − 0.01 = 0
Solve the quadratic
x = (−0.1 + √(0.1² + 4·0.01))/2 = 0.0618 mol/L
Equilibrium concentration
[Cl₂] = x = 0.0618 mol/L
Answer
[Cl2] = 0.0618 mol/L
11.pH & buffers
Answer: pH = 2.759
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/(0.05 − x) → x = 0.001744 mol/L
pH
pH = −log(0.001744) = 2.759
Answer
pH = 2.759
12.pH & buffers
Answer: pH = 4.046
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(0.1/0.5) = 4.046
Answer
pH = 4.046
13.Reaction kinetics
Answer: t = 3.054 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 = 0.4
Solve for the time
t = −ln(1 − X)/k = −ln(0.4)/0.3 = 3.054 min
Answer
t = 3.054 min
14.Reaction kinetics
Answer: k2 = 0.3568 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).