4 g of Na and 24 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 2Mg + O2 → 2MgO, 20 g of Mg reacts completely (the other reactants are in excess) and 31.17 g of MgO is collected. Find the percent yield Y.
M(Mg) = 24.31 g/molM(MgO) = 40.3 g/molactual: 31.17 g
Y = %
3.Molarity & dilutionMedium
You need 1000 mL of 0.2 mol/L solution and have a 0.5 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 0.5 mol/Lc2 = 0.2 mol/LV2 = 1000 mL
V1 = mL
4.Molarity & dilutionHard
150 mL of a 0.8 mol/L solution is mixed with 500 mL of a 0.1 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 75 g block of aluminium (c = 0.897 J/(g·°C)) at 90 °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.
Use the standard enthalpies of formation to find the heat q released when 250 g of C2H5OH (M = 46.07 g/mol) burns completely: C2H5OH + 3O₂ → 2CO₂ + 3H₂O(l). Give q as a positive number.
0.4 mol/L of H₂ and 0.4 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 = 0.4 mol/LKc = 9
[HI] = 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 0.2 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).
c = 0.2 mol/LKa = 6.3 × 10⁻⁵
pH =
12.pH & buffersHard
A buffer contains 0.1 mol/L lactic acid and 0.3 mol/L of its sodium salt. Ka = 0.00014. Find the pH pH of the buffer.
[HA] = 0.1[A⁻] = 0.3Ka = 0.00014
pH =
13.Reaction kineticsMedium
A first-order reaction in a batch reactor has k = 0.15 min⁻¹. How long does it take to reach 99% conversion? Find t.
k = 0.15 min⁻¹X = 99 %
t = min
14.Reaction kineticsHard
A reaction has an activation energy of 40 kJ/mol. At 30 °C it is too slow. To what temperature T2 (in °C) must it be heated to make it 10 times faster?
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) = 4/22.99 = 0.174 mol
n(Cl2) = 24/70.9 = 0.3385 mol
Limiting reactant
0.174/2 = 0.08699, 0.3385/1 = 0.3385 → Na
Mass of product
n(NaCl) = 0.08699·2 = 0.174 mol
m = 0.174·58.44 = 10.17 g
Answer
m = 10.17 g
2.Stoichiometry
Answer: Y = 94.01 %
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(Mg) = 20/24.31 = 0.8227 mol
m = 0.8227·2/2·40.3 = 33.16 g
Percent yield
Y = 31.17/33.16·100 = 94.01%
Answer
Y = 94.01%
3.Molarity & dilution
Answer: V1 = 400 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 = 0.2·1000/0.5 = 400 mL
Answer
V1 = 400 mL
4.Molarity & dilution
Answer: c = 0.2615 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 = 0.8·150 + 0.1·500 = 170 mmol
Concentration of the mixture
c = 170/650 = 0.2615 mol/L
Answer
c = 0.2615 mol/L
5.Ideal gas law
Answer: V = 18.86 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 170/39.95 = 4.255 mol
T = 20 + 273.15 = 293.1 K
Volume
V = nRT/P = 4.255·8.314·293.1/550 = 18.86 L
Answer
V = 18.86 L
6.Ideal gas law
Answer: ρ = 6.889 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 90 + 273.15 = 363.1 K
Density
ρ = 650·32/(8.314·363.1) = 6.889 kg/m³
Answer
ρ = 6.889 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 21.42 °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
75·0.897·(90 − Tf) = 250·4.18·(Tf − 17)
Solve for the final temperature
Tf = (67.28·90 + 1045·17)/(67.28 + 1045) = 21.42 °C
Answer
Tf = 21.42 °C
8.Calorimetry & enthalpy
Answer: q = 7416 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.
Equilibrium expression
9 = (2x)²/(0.4 − x)²
Take the square root
3 = 2x/(0.4 − x) → x = 0.4·3/(2 + 3) = 0.24 mol/L
Equilibrium concentration
[HI] = 2x = 0.48 mol/L
Answer
[HI] = 0.48 mol/L
10.Chemical equilibrium
Answer: [PCl5] = 0.3416 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
0.01 = 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 = 2.454
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/(0.2 − x) → x = 0.003518 mol/L
pH
pH = −log(0.003518) = 2.454
Answer
pH = 2.454
12.pH & buffers
Answer: pH = 4.331
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(0.3/0.1) = 4.331
Answer
pH = 4.331
13.Reaction kinetics
Answer: t = 30.7 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 = 0.01
Solve for the time
t = −ln(1 − X)/k = −ln(0.01)/0.15 = 30.7 min
Answer
t = 30.7 min
14.Reaction kinetics
Answer: T2 = 81.45 °C
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).