4 g of Al and 20 g of Cl2 react according to 2Al + 3Cl2 → 2AlCl3. What mass m of AlCl3 can form?
M(Al) = 26.98M(Cl2) = 70.9M(AlCl3) = 133.33 g/mol
m = g
2.StoichiometryHard
In the reaction CaCO3 → CaO + CO2, 30 g of CaCO3 reacts completely (the other reactants are in excess) and 9.23 g of CO2 is collected. Find the percent yield Y.
M(CaCO3) = 100.09 g/molM(CO2) = 44.01 g/molactual: 9.23 g
Y = %
3.Molarity & dilutionMedium
You need 100 mL of 0.1 mol/L solution and have a 5 mol/L stock solution. How much water Vwater do you add to the stock solution?
c1 = 5 mol/Lc2 = 0.1 mol/LV2 = 100 mL
Vwater = mL
4.Molarity & dilutionHard
300 mL of a 0.8 mol/L solution is mixed with 400 mL of a 0.3 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 75 g block of iron (c = 0.449 J/(g·°C)) at 130 °C is dropped into 200 g of water (c = 4.18 J/(g·°C)) at 24 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find the heat q released when 10 g of C6H12O6 (M = 180.16 g/mol) burns completely: C6H12O6 + 6O₂ → 6CO₂ + 6H₂O(l). Give q as a positive number.
0.25 mol/L of H₂ and 0.25 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 9. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.25 mol/LKc = 9
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 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 = 0.08
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.25 mol/L solution of lactic acid (Ka = 0.00014).
c = 0.25 mol/LKa = 0.00014
pH =
12.pH & buffersHard
A buffer contains 0.5 mol/L lactic acid and 0.5 mol/L of its sodium salt. Ka = 0.00014. Find the pH pH of the buffer.
[HA] = 0.5[A⁻] = 0.5Ka = 0.00014
pH =
13.Reaction kineticsMedium
A second-order reaction (−rA = k·CA²) in a batch reactor has k = 0.8 L/(mol·min) and starts at 1.5 mol/L. How long does it take to reach 80% conversion? Find t.
k = 0.8 L/(mol·min)CA0 = 1.5 mol/LX = 80 %
t = min
14.Reaction kineticsHard
The rate constant of a reaction is 0.1 min⁻¹ at 35 °C and 0.851 min⁻¹ at 55 °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(Al) = 4/26.98 = 0.1483 mol
n(Cl2) = 20/70.9 = 0.2821 mol
Limiting reactant
0.1483/2 = 0.07413, 0.2821/3 = 0.09403 → Al
Mass of product
n(AlCl3) = 0.07413·2 = 0.1483 mol
m = 0.1483·133.33 = 19.77 g
Answer
m = 19.77 g
2.Stoichiometry
Answer: Y = 69.97 %
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) = 30/100.09 = 0.2997 mol
m = 0.2997·1/1·44.01 = 13.19 g
Percent yield
Y = 9.23/13.19·100 = 69.97%
Answer
Y = 69.97%
3.Molarity & dilution
Answer: Vwater = 98 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 = 0.1·100/5 = 2 mL
Water to add
Vwater = 100 − 2 = 98 mL
Answer
Vwater = 98 mL
4.Molarity & dilution
Answer: c = 0.5143 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 = 0.8·300 + 0.3·400 = 360 mmol
Concentration of the mixture
c = 360/700 = 0.5143 mol/L
Answer
c = 0.5143 mol/L
5.Ideal gas law
Answer: V = 28.27 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 60/16.04 = 3.741 mol
T = 45 + 273.15 = 318.1 K
Volume
V = nRT/P = 3.741·8.314·318.1/350 = 28.27 L
Answer
V = 28.27 L
6.Ideal gas law
Answer: ρ = 3.365 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 70 + 273.15 = 343.1 K
Density
ρ = 300·32/(8.314·343.1) = 3.365 kg/m³
Answer
ρ = 3.365 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 28.1 °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
75·0.449·(130 − Tf) = 200·4.18·(Tf − 24)
Solve for the final temperature
Tf = (33.68·130 + 836·24)/(33.68 + 836) = 28.1 °C
Answer
Tf = 28.1 °C
8.Calorimetry & enthalpy
Answer: q = 155.5 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
0.08 = x²/(0.8 − x) → x² + 0.08·x − 0.064 = 0
Solve the quadratic
x = (−0.08 + √(0.08² + 4·0.064))/2 = 0.2161 mol/L
Equilibrium concentration
[Cl₂] = x = 0.2161 mol/L
Answer
[Cl2] = 0.2161 mol/L
11.pH & buffers
Answer: pH = 2.233
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 = 0.005846 mol/L
pH
pH = −log(0.005846) = 2.233
Answer
pH = 2.233
12.pH & buffers
Answer: pH = 3.854
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(0.5/0.5) = 3.854
Answer
pH = 3.854
13.Reaction kinetics
Answer: t = 3.333 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).
Concentration at the end
CA = 1.5·(1 − 0.8) = 0.3 mol/L
Integrated second-order law
t = (1/CA − 1/CA0)/k = (3.333 − 0.6667)/0.8 = 3.333 min
Answer
t = 3.333 min
14.Reaction kinetics
Answer: Ea = 90.01 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 = 308.1 K, T2 = 328.1 K
Solve Arrhenius for Ea
Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·2.141/(-1.978e-4) = 90008 J/mol