10 g of N2 and 10 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 2Na + Cl2 → 2NaCl, 12 g of Na reacts completely (the other reactants are in excess) and 26.84 g of NaCl is collected. Find the percent yield Y.
M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 26.84 g
Y = %
3.Molarity & dilutionMedium
You need 1000 mL of 0.5 mol/L solution and have a 3 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 3 mol/Lc2 = 0.5 mol/LV2 = 1000 mL
V1 = mL
4.Molarity & dilutionHard
200 mL of a 0.25 mol/L solution is mixed with 250 mL of a 2.5 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 200 g block of lead (c = 0.129 J/(g·°C)) at 80 °C is dropped into 300 g of water (c = 4.18 J/(g·°C)) at 16 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find the heat q released when 20 g of C2H6 (M = 30.07 g/mol) burns completely: 2C2H6 + 7O₂ → 4CO₂ + 6H₂O(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 = 50. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.5 mol/LKc = 50
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 at the reaction temperature. A container starts with 0.5 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 0.5 mol/LKc = 0.08
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.5 mol/L solution of lactic acid (Ka = 0.00014).
c = 0.5 mol/LKa = 0.00014
pH =
12.pH & buffersHard
A 1.00 L buffer contains 0.3 mol benzoic acid and 0.1 mol of its conjugate base (Ka = 6.3 × 10⁻⁵). 0.05 mol of NaOH is added. Find the new pH pH.
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) = 10/28.02 = 0.3569 mol
n(H2) = 10/2.016 = 4.96 mol
Limiting reactant
0.3569/1 = 0.3569, 4.96/3 = 1.653 → N2
Mass of product
n(NH3) = 0.3569·2 = 0.7138 mol
m = 0.7138·17.03 = 12.16 g
Answer
m = 12.16 g
2.Stoichiometry
Answer: Y = 87.99 %
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(Na) = 12/22.99 = 0.522 mol
m = 0.522·2/2·58.44 = 30.5 g
Percent yield
Y = 26.84/30.5·100 = 87.99%
Answer
Y = 87.99%
3.Molarity & dilution
Answer: V1 = 166.7 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 = 0.5·1000/3 = 166.7 mL
Answer
V1 = 166.7 mL
4.Molarity & dilution
Answer: c = 1.5 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 + 2.5·250 = 675 mmol
Concentration of the mixture
c = 675/450 = 1.5 mol/L
Answer
c = 1.5 mol/L
5.Ideal gas law
Answer: P = 175.2 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 100/44.01 = 2.272 mol
T = 5 + 273.15 = 278.1 K
Pressure
P = nRT/V = 2.272·8.314·278.1/0.03 = 175.2 kPa
Answer
P = 175.2 kPa
6.Ideal gas law
Answer: V2 = 10.28 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Absolute temperatures
T1 = 298.1 K, T2 = 383.1 K
Combined gas law
V2 = V1·(P1/P2)·(T2/T1) = 10·(400/500)·(383.1/298.1) = 10.28 L
Answer
V2 = 10.28 L
7.Calorimetry & enthalpy
Answer: Tf = 17.29 °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
200·0.129·(80 − Tf) = 300·4.18·(Tf − 16)
Solve for the final temperature
Tf = (25.8·80 + 1254·16)/(25.8 + 1254) = 17.29 °C
Answer
Tf = 17.29 °C
8.Calorimetry & enthalpy
Answer: q = 1037 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/(0.5 − x) → x² + 0.08·x − 0.04 = 0
Solve the quadratic
x = (−0.08 + √(0.08² + 4·0.04))/2 = 0.164 mol/L
Equilibrium concentration
[Cl₂] = x = 0.164 mol/L
Answer
[Cl2] = 0.164 mol/L
11.pH & buffers
Answer: pH = 2.081
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/(0.5 − x) → x = 0.008297 mol/L
pH
pH = −log(0.008297) = 2.081
Answer
pH = 2.081
12.pH & buffers
Answer: pH = 3.979
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]).
Neutralisation
n(HA) = 0.3 − 0.05 = 0.25 mol
n(A⁻) = 0.1 + 0.05 = 0.15 mol
Henderson–Hasselbalch
pH = 4.201 + log(0.15/0.25) = 3.979
Answer
pH = 3.979
13.Reaction kinetics
Answer: t = 69.31 min
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).
Conversion and concentration
C/C0 = 1 − X = 0.25
Solve for the time
t = −ln(1 − X)/k = −ln(0.25)/0.02 = 69.31 min
Answer
t = 69.31 min
14.Reaction kinetics
Answer: k2 = 0.1252 min⁻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).