6 g of C3H8 and 80 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of CO2 can form?
M(C3H8) = 44.1M(O2) = 32M(CO2) = 44.01 g/mol
m = g
2.StoichiometryHard
In the reaction CaCO3 → CaO + CO2, 10 g of CaCO3 reacts completely (the other reactants are in excess) and 3.87 g of CO2 is collected. Find the percent yield Y.
M(CaCO3) = 100.09 g/molM(CO2) = 44.01 g/molactual: 3.87 g
Y = %
3.Molarity & dilutionMedium
You need 100 mL of 0.25 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 = 0.25 mol/LV2 = 100 mL
Vwater = mL
4.Molarity & dilutionHard
200 mL of a 0.25 mol/L solution is mixed with 50 mL of a 0.05 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 150 g block of zinc (c = 0.388 J/(g·°C)) at 90 °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 50 g of C3H8 (M = 44.1 g/mol) burns completely: C3H8 + 5O₂ → 3CO₂ + 4H₂O(l). Give q as a positive number.
0.5 mol/L of H₂ and 0.5 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 25. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.5 mol/LKc = 25
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.5 at the reaction temperature. A container starts with 1.5 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 1.5 mol/LKc = 0.5
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.01 mol/L solution of formic acid (Ka = 0.00018).
c = 0.01 mol/LKa = 0.00018
pH =
12.pH & buffersHard
A buffer contains 0.5 mol/L hypochlorous acid and 0.2 mol/L of its sodium salt. Ka = 3 × 10⁻⁸. Find the pH pH of the buffer.
[HA] = 0.5[A⁻] = 0.2Ka = 3 × 10⁻⁸
pH =
13.Reaction kineticsMedium
A zero-order reaction has k = 0.05 mol/(L·min) and starts at 2 mol/L. How long does it take to reach 20% conversion? Find t.
k = 0.05 mol/(L·min)CA0 = 2 mol/LX = 20 %
t = min
14.Reaction kineticsHard
The rate constant of a reaction is 0.1 min⁻¹ at 40 °C and 0.564 min⁻¹ at 60 °C. Find the activation energy Ea.
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(C3H8) = 6/44.1 = 0.1361 mol
n(O2) = 80/32 = 2.5 mol
Limiting reactant
0.1361/1 = 0.1361, 2.5/5 = 0.5 → C3H8
Mass of product
n(CO2) = 0.1361·3 = 0.4082 mol
m = 0.4082·44.01 = 17.96 g
Answer
m = 17.96 g
2.Stoichiometry
Answer: Y = 88.01 %
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(CaCO3) = 10/100.09 = 0.09991 mol
m = 0.09991·1/1·44.01 = 4.397 g
Percent yield
Y = 3.87/4.397·100 = 88.01%
Answer
Y = 88.01%
3.Molarity & dilution
Answer: Vwater = 93.75 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 = 0.25·100/4 = 6.25 mL
Water to add
Vwater = 100 − 6.25 = 93.75 mL
Answer
Vwater = 93.75 mL
4.Molarity & dilution
Answer: c = 0.21 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 = 0.25·200 + 0.05·50 = 52.5 mmol
Concentration of the mixture
c = 52.5/250 = 0.21 mol/L
Answer
c = 0.21 mol/L
5.Ideal gas law
Answer: V = 32.38 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 320/44.01 = 7.271 mol
T = 75 + 273.15 = 348.1 K
Volume
V = nRT/P = 7.271·8.314·348.1/650 = 32.38 L
Answer
V = 32.38 L
6.Ideal gas law
Answer: ρ = 3.546 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 100 + 273.15 = 373.1 K
Density
ρ = 250·44.01/(8.314·373.1) = 3.546 kg/m³
Answer
ρ = 3.546 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 24.64 °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
150·0.388·(90 − Tf) = 250·4.18·(Tf − 21)
Solve for the final temperature
Tf = (58.2·90 + 1045·21)/(58.2 + 1045) = 24.64 °C
Answer
Tf = 24.64 °C
8.Calorimetry & enthalpy
Answer: q = 2517 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
0.5 = x²/(1.5 − x) → x² + 0.5·x − 0.75 = 0
Solve the quadratic
x = (−0.5 + √(0.5² + 4·0.75))/2 = 0.6514 mol/L
Equilibrium concentration
[Cl₂] = x = 0.6514 mol/L
Answer
[Cl2] = 0.6514 mol/L
11.pH & buffers
Answer: pH = 2.901
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 = 0.001255 mol/L
pH
pH = −log(0.001255) = 2.901
Answer
pH = 2.901
12.pH & buffers
Answer: pH = 7.125
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 = 7.523
Henderson–Hasselbalch
pH = 7.523 + log(0.2/0.5) = 7.125
Answer
pH = 7.125
13.Reaction kinetics
Answer: t = 8 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 = 2·0.2 = 0.4 mol/L
Time
t = CA0·X/k = 0.4/0.05 = 8 min
Answer
t = 8 min
14.Reaction kinetics
Answer: Ea = 75.02 kJ/mol
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).
Absolute temperatures
T1 = 313.1 K, T2 = 333.1 K
Solve Arrhenius for Ea
Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·1.73/(-1.917e-4) = 75022 J/mol