10 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 CaCO3 → CaO + CO2, 24 g of CaCO3 reacts completely (the other reactants are in excess) and 8.23 g of CO2 is collected. Find the percent yield Y.
M(CaCO3) = 100.09 g/molM(CO2) = 44.01 g/molactual: 8.23 g
Y = %
3.Molarity & dilutionMedium
You need 200 mL of 0.25 mol/L solution and have a 5 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 5 mol/Lc2 = 0.25 mol/LV2 = 200 mL
V1 = mL
4.Molarity & dilutionHard
150 mL of a 0.25 mol/L solution is mixed with 400 mL of a 1.2 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 250 g block of copper (c = 0.385 J/(g·°C)) at 120 °C is dropped into 150 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 50 g of CH4 (M = 16.04 g/mol) burns completely: CH4 + 2O₂ → CO₂ + 2H₂O(l). Give q as a positive number.
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 = 0.05 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 = 0.05
[PCl5] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.25 mol/L solution of formic acid (Ka = 0.00018).
c = 0.25 mol/LKa = 0.00018
pH =
12.pH & buffersHard
A buffer contains 0.4 mol/L acetic acid and 0.2 mol/L of its sodium salt. Ka = 1.8 × 10⁻⁵. Find the pH pH of the buffer.
[HA] = 0.4[A⁻] = 0.2Ka = 1.8 × 10⁻⁵
pH =
13.Reaction kineticsMedium
A reaction with rate law −rA = k·CA² has k = 0.8 L/(mol·min) and starts at 2 mol/L. Find the concentration CA after 30 min.
k = 0.8 L/(mol·min)CA0 = 2 mol/Lt = 30 min
CA = mol/L
14.Reaction kineticsHard
A reaction has an activation energy of 100 kJ/mol. At 25 °C it is too slow. To what temperature T2 (in °C) must it be heated to make it 5 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) = 10/22.99 = 0.435 mol
n(Cl2) = 24/70.9 = 0.3385 mol
Limiting reactant
0.435/2 = 0.2175, 0.3385/1 = 0.3385 → Na
Mass of product
n(NaCl) = 0.2175·2 = 0.435 mol
m = 0.435·58.44 = 25.42 g
Answer
m = 25.42 g
2.Stoichiometry
Answer: Y = 77.99 %
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(CaCO3) = 24/100.09 = 0.2398 mol
m = 0.2398·1/1·44.01 = 10.55 g
Percent yield
Y = 8.23/10.55·100 = 77.99%
Answer
Y = 77.99%
3.Molarity & dilution
Answer: V1 = 10 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 = 0.25·200/5 = 10 mL
Answer
V1 = 10 mL
4.Molarity & dilution
Answer: c = 0.9409 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 = 0.25·150 + 1.2·400 = 517.5 mmol
Concentration of the mixture
c = 517.5/550 = 0.9409 mol/L
Answer
c = 0.9409 mol/L
5.Ideal gas law
Answer: V = 46.48 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 420/32 = 13.13 mol
T = 25 + 273.15 = 298.1 K
Volume
V = nRT/P = 13.13·8.314·298.1/700 = 46.48 L
Answer
V = 46.48 L
6.Ideal gas law
Answer: ρ = 2.915 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 = 200·44.01/(8.314·363.1) = 2.915 kg/m³
Answer
ρ = 2.915 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 31.57 °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
250·0.385·(120 − Tf) = 150·4.18·(Tf − 18)
Solve for the final temperature
Tf = (96.25·120 + 627·18)/(96.25 + 627) = 31.57 °C
Answer
Tf = 31.57 °C
8.Calorimetry & enthalpy
Answer: q = 2775 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
0.05 = x²/(1.5 − x) → x² + 0.05·x − 0.075 = 0
Solve the quadratic
x = (−0.05 + √(0.05² + 4·0.075))/2 = 0.25 mol/L
Equilibrium concentration
[PCl₅] = 1.5 − 0.25 = 1.25 mol/L
Answer
[PCl5] = 1.25 mol/L
11.pH & buffers
Answer: pH = 2.179
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/(0.25 − x) → x = 0.006619 mol/L
pH
pH = −log(0.006619) = 2.179
Answer
pH = 2.179
12.pH & buffers
Answer: pH = 4.444
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(0.2/0.4) = 4.444
Answer
pH = 4.444
13.Reaction kinetics
Answer: CA = 0.04082 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 = 0.5 + 0.8·30 = 24.5 L/mol
CA = 1/24.5 = 0.04082 mol/L
Answer
CA = 0.04082 mol/L
14.Reaction kinetics
Answer: T2 = 37.39 °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).