30 g of CH4 and 40 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, 20 g of Al reacts completely (the other reactants are in excess) and 86.98 g of AlCl3 is collected. Find the percent yield Y.
M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 86.98 g
Y = %
3.Molarity & dilutionMedium
You need 250 mL of 0.4 mol/L solution and have a 1 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 1 mol/Lc2 = 0.4 mol/LV2 = 250 mL
V1 = mL
4.Molarity & dilutionHard
300 mL of a 0.5 mol/L solution is mixed with 250 mL of a 0.1 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 200 g block of aluminium (c = 0.897 J/(g·°C)) at 130 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 17 °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 = 9. Find the equilibrium concentration [HI].
[H₂]₀ = [I₂]₀ = 0.2 mol/LKc = 9
[HI] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.05 at the reaction temperature. A container starts with 0.8 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 0.8 mol/LKc = 0.05
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.01 mol/L solution of hypochlorous acid (Ka = 3 × 10⁻⁸).
c = 0.01 mol/LKa = 3 × 10⁻⁸
pH =
12.pH & buffersHard
A buffer contains 0.4 mol/L hydrofluoric acid and 0.2 mol/L of its sodium salt. Ka = 0.00068. Find the pH pH of the buffer.
[HA] = 0.4[A⁻] = 0.2Ka = 0.00068
pH =
13.Reaction kineticsMedium
A zero-order reaction has k = 0.005 mol/(L·min) and starts at 1.5 mol/L. How long does it take to reach 75% conversion? Find t.
k = 0.005 mol/(L·min)CA0 = 1.5 mol/LX = 75 %
t = min
14.Reaction kineticsHard
A reaction has a rate constant of 0.05 min⁻¹ at 20 °C and an activation energy of 90 kJ/mol. Find the rate constant k2 at 35 °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(CH4) = 30/16.04 = 1.87 mol
n(O2) = 40/32 = 1.25 mol
Limiting reactant
1.87/1 = 1.87, 1.25/2 = 0.625 → O2
Mass of product
n(CO2) = 0.625·1 = 0.625 mol
m = 0.625·44.01 = 27.51 g
Answer
m = 27.51 g
2.Stoichiometry
Answer: Y = 88 %
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) = 20/26.98 = 0.7413 mol
m = 0.7413·2/2·133.33 = 98.84 g
Percent yield
Y = 86.98/98.84·100 = 88%
Answer
Y = 88%
3.Molarity & dilution
Answer: V1 = 100 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 = 0.4·250/1 = 100 mL
Answer
V1 = 100 mL
4.Molarity & dilution
Answer: c = 0.3182 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 = 0.5·300 + 0.1·250 = 175 mmol
Concentration of the mixture
c = 175/550 = 0.3182 mol/L
Answer
c = 0.3182 mol/L
5.Ideal gas law
Answer: P = 606.2 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 340/44.01 = 7.726 mol
T = 10 + 273.15 = 283.1 K
Pressure
P = nRT/V = 7.726·8.314·283.1/0.03 = 606.2 kPa
Answer
P = 606.2 kPa
6.Ideal gas law
Answer: ρ = 2.91 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 190 + 273.15 = 463.1 K
Density
ρ = 400·28.01/(8.314·463.1) = 2.91 kg/m³
Answer
ρ = 2.91 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 33.56 °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
200·0.897·(130 − Tf) = 250·4.18·(Tf − 17)
Solve for the final temperature
Tf = (179.4·130 + 1045·17)/(179.4 + 1045) = 33.56 °C
Answer
Tf = 33.56 °C
8.Calorimetry & enthalpy
Answer: ΔH° = −1453 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] = 0.24 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
9 = (2x)²/(0.2 − x)²
Take the square root
3 = 2x/(0.2 − x) → x = 0.2·3/(2 + 3) = 0.12 mol/L
Equilibrium concentration
[HI] = 2x = 0.24 mol/L
Answer
[HI] = 0.24 mol/L
10.Chemical equilibrium
Answer: [Cl2] = 0.1766 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
0.05 = x²/(0.8 − x) → x² + 0.05·x − 0.04 = 0
Solve the quadratic
x = (−0.05 + √(0.05² + 4·0.04))/2 = 0.1766 mol/L
Equilibrium concentration
[Cl₂] = x = 0.1766 mol/L
Answer
[Cl2] = 0.1766 mol/L
11.pH & buffers
Answer: pH = 4.762
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 = 1.731 × 10⁻⁵ mol/L
pH
pH = −log(1.731 × 10⁻⁵) = 4.762
Answer
pH = 4.762
12.pH & buffers
Answer: pH = 2.866
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 = 3.167
Henderson–Hasselbalch
pH = 3.167 + log(0.2/0.4) = 2.866
Answer
pH = 2.866
13.Reaction kinetics
Answer: t = 225 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).
Amount converted
CA0·X = 1.5·0.75 = 1.125 mol/L
Time
t = CA0·X/k = 1.125/0.005 = 225 min
Answer
t = 225 min
14.Reaction kinetics
Answer: k2 = 0.3017 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).