4 g of H2 and 8 g of O2 react according to 2H2 + O2 → 2H2O. What mass m of H2O can form?
M(H2) = 2.016M(O2) = 32M(H2O) = 18.02 g/mol
m = g
2.StoichiometryHard
In the reaction N2 + 3H2 → 2NH3, 16 g of N2 reacts completely (the other reactants are in excess) and 15.17 g of NH3 is collected. Find the percent yield Y.
M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 15.17 g
Y = %
3.Molarity & dilutionMedium
You need 200 mL of 0.05 mol/L solution and have a 4 mol/L stock solution. How much water Vwater do you add to the stock solution?
c1 = 4 mol/Lc2 = 0.05 mol/LV2 = 200 mL
Vwater = mL
4.Molarity & dilutionHard
250 mL of a 1 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 = 1 mol/LV1 = 250 mLc2 = 0.75 mol/LV2 = 100 mL
c = mol/L
5.Ideal gas lawMedium
A tank of 5 L contains 340 g of He at 70 °C. Treat it as an ideal gas and find the pressure P.
A 150 g block of iron (c = 0.449 J/(g·°C)) at 150 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 21 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find the heat q released when 25 g of C2H6 (M = 30.07 g/mol) burns completely: 2C2H6 + 7O₂ → 4CO₂ + 6H₂O(l). Give q as a positive number.
0.1 mol/L of H₂ and 0.1 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 4. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.1 mol/LKc = 4
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.1 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 = 0.1
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.05 mol/L solution of lactic acid (Ka = 0.00014).
c = 0.05 mol/LKa = 0.00014
pH =
12.pH & buffersHard
A buffer contains 0.25 mol/L hydrofluoric acid and 0.3 mol/L of its sodium salt. Ka = 0.00068. Find the pH pH of the buffer.
[HA] = 0.25[A⁻] = 0.3Ka = 0.00068
pH =
13.Reaction kineticsMedium
A reaction with rate law −rA = k·CA² has k = 0.2 L/(mol·min) and starts at 1 mol/L. Find the concentration CA after 15 min.
k = 0.2 L/(mol·min)CA0 = 1 mol/Lt = 15 min
CA = mol/L
14.Reaction kineticsHard
A reaction has an activation energy of 70 kJ/mol. At 25 °C it is too slow. To what temperature T2 (in °C) must it be heated to make it 3 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(H2) = 4/2.016 = 1.984 mol
n(O2) = 8/32 = 0.25 mol
Limiting reactant
1.984/2 = 0.9921, 0.25/1 = 0.25 → O2
Mass of product
n(H2O) = 0.25·2 = 0.5 mol
m = 0.5·18.02 = 9.01 g
Answer
m = 9.01 g
2.Stoichiometry
Answer: Y = 78 %
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(N2) = 16/28.02 = 0.571 mol
m = 0.571·2/1·17.03 = 19.45 g
Percent yield
Y = 15.17/19.45·100 = 78%
Answer
Y = 78%
3.Molarity & dilution
Answer: Vwater = 197.5 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 = 0.05·200/4 = 2.5 mL
Water to add
Vwater = 200 − 2.5 = 197.5 mL
Answer
Vwater = 197.5 mL
4.Molarity & dilution
Answer: c = 0.9286 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 = 1·250 + 0.75·100 = 325 mmol
Concentration of the mixture
c = 325/350 = 0.9286 mol/L
Answer
c = 0.9286 mol/L
5.Ideal gas law
Answer: P = 48464 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 340/4.003 = 84.94 mol
T = 70 + 273.15 = 343.1 K
Pressure
P = nRT/V = 84.94·8.314·343.1/0.005 = 48464 kPa
Answer
P = 48464 kPa
6.Ideal gas law
Answer: ρ = 9.847 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 20 + 273.15 = 293.1 K
Density
ρ = 750·32/(8.314·293.1) = 9.847 kg/m³
Answer
ρ = 9.847 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 28.81 °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
150·0.449·(150 − Tf) = 250·4.18·(Tf − 21)
Solve for the final temperature
Tf = (67.35·150 + 1045·21)/(67.35 + 1045) = 28.81 °C
Answer
Tf = 28.81 °C
8.Calorimetry & enthalpy
Answer: q = 1297 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
4 = (2x)²/(0.1 − x)²
Take the square root
2 = 2x/(0.1 − x) → x = 0.1·2/(2 + 2) = 0.05 mol/L
Equilibrium concentration
[H₂] = 0.1 − x = 0.05 mol/L
Answer
[H2] = 0.05 mol/L
10.Chemical equilibrium
Answer: [Cl2] = 0.2372 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
0.1 = x²/(0.8 − x) → x² + 0.1·x − 0.08 = 0
Solve the quadratic
x = (−0.1 + √(0.1² + 4·0.08))/2 = 0.2372 mol/L
Equilibrium concentration
[Cl₂] = x = 0.2372 mol/L
Answer
[Cl2] = 0.2372 mol/L
11.pH & buffers
Answer: pH = 2.589
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 = 0.002577 mol/L
pH
pH = −log(0.002577) = 2.589
Answer
pH = 2.589
12.pH & buffers
Answer: pH = 3.247
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(0.3/0.25) = 3.247
Answer
pH = 3.247
13.Reaction kinetics
Answer: CA = 0.25 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 = 1 + 0.2·15 = 4 L/mol
CA = 1/4 = 0.25 mol/L
Answer
CA = 0.25 mol/L
14.Reaction kinetics
Answer: T2 = 37.07 °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).