12 g of N2 and 32 g of H2 react according to N2 + 3H2 → 2NH3. What mass m of NH3 can form?
M(N2) = 28.02M(H2) = 2.016M(NH3) = 17.03 g/mol
m = g
2.StoichiometryHard
In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 24 g of Fe2O3 reacts completely (the other reactants are in excess) and 12.3 g of CO2 is collected. Find the percent yield Y.
M(Fe2O3) = 159.69 g/molM(CO2) = 44.01 g/molactual: 12.3 g
Y = %
3.Molarity & dilutionMedium
You need 200 mL of 0.05 mol/L solution and have a 3 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 3 mol/Lc2 = 0.05 mol/LV2 = 200 mL
V1 = mL
4.Molarity & dilutionHard
100 mL of a 0.5 mol/L solution is mixed with 250 mL of a 0.75 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 100 g block of aluminium (c = 0.897 J/(g·°C)) at 100 °C is dropped into 100 g of water (c = 4.18 J/(g·°C)) at 23 °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 CH4 (M = 16.04 g/mol) burns completely: CH4 + 2O₂ → CO₂ + 2H₂O(l). Give q as a positive number.
0.2 mol/L of H₂ and 0.2 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 49. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.2 mol/LKc = 49
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 at the reaction temperature. A container starts with 0.4 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 0.4 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 benzoic acid and 0.1 mol/L of its sodium salt. Ka = 6.3 × 10⁻⁵. Find the pH pH of the buffer.
[HA] = 0.5[A⁻] = 0.1Ka = 6.3 × 10⁻⁵
pH =
13.Reaction kineticsMedium
A second-order reaction (−rA = k·CA²) in a batch reactor has k = 0.5 L/(mol·min) and starts at 1 mol/L. How long does it take to reach 75% conversion? Find t.
k = 0.5 L/(mol·min)CA0 = 1 mol/LX = 75 %
t = min
14.Reaction kineticsHard
A first-order reaction has a rate constant of 0.05 s⁻¹ at 100 °C and an activation energy of 60 kJ/mol. Find the pre-exponential factor A in the Arrhenius equation.
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(N2) = 12/28.02 = 0.4283 mol
n(H2) = 32/2.016 = 15.87 mol
Limiting reactant
0.4283/1 = 0.4283, 15.87/3 = 5.291 → N2
Mass of product
n(NH3) = 0.4283·2 = 0.8565 mol
m = 0.8565·17.03 = 14.59 g
Answer
m = 14.59 g
2.Stoichiometry
Answer: Y = 61.99 %
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(Fe2O3) = 24/159.69 = 0.1503 mol
m = 0.1503·3/1·44.01 = 19.84 g
Percent yield
Y = 12.3/19.84·100 = 61.99%
Answer
Y = 61.99%
3.Molarity & dilution
Answer: V1 = 3.333 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 = 0.05·200/3 = 3.333 mL
Answer
V1 = 3.333 mL
4.Molarity & dilution
Answer: c = 0.6786 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 = 0.5·100 + 0.75·250 = 237.5 mmol
Concentration of the mixture
c = 237.5/350 = 0.6786 mol/L
Answer
c = 0.6786 mol/L
5.Ideal gas law
Answer: P = 2396 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 320/32 = 10 mol
T = 15 + 273.15 = 288.1 K
Pressure
P = nRT/V = 10·8.314·288.1/0.01 = 2396 kPa
Answer
P = 2396 kPa
6.Ideal gas law
Answer: ρ = 1.229 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 40 + 273.15 = 313.1 K
Density
ρ = 100·32/(8.314·313.1) = 1.229 kg/m³
Answer
ρ = 1.229 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 36.6 °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
100·0.897·(100 − Tf) = 100·4.18·(Tf − 23)
Solve for the final temperature
Tf = (89.7·100 + 418·23)/(89.7 + 418) = 36.6 °C
Answer
Tf = 36.6 °C
8.Calorimetry & enthalpy
Answer: q = 13876 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/(0.4 − x) → x² + 0.08·x − 0.032 = 0
Solve the quadratic
x = (−0.08 + √(0.08² + 4·0.032))/2 = 0.1433 mol/L
Equilibrium concentration
[Cl₂] = x = 0.1433 mol/L
Answer
[Cl2] = 0.1433 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 = 3.502
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 = 4.201
Henderson–Hasselbalch
pH = 4.201 + log(0.1/0.5) = 3.502
Answer
pH = 3.502
13.Reaction kinetics
Answer: t = 6 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·(1 − 0.75) = 0.25 mol/L
Integrated second-order law
t = (1/CA − 1/CA0)/k = (4 − 1)/0.5 = 6 min
Answer
t = 6 min
14.Reaction kinetics
Answer: A = 1.254 × 10⁷ s⁻¹
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).