20 g of Na and 8 g of Cl2 react according to 2Na + Cl2 → 2NaCl. What mass m of NaCl can form?
M(Na) = 22.99M(Cl2) = 70.9M(NaCl) = 58.44 g/mol
m = g
2.StoichiometryHard
In the reaction 2Mg + O2 → 2MgO, 30 g of Mg reacts completely (the other reactants are in excess) and 34.81 g of MgO is collected. Find the percent yield Y.
M(Mg) = 24.31 g/molM(MgO) = 40.3 g/molactual: 34.81 g
Y = %
3.Molarity & dilutionMedium
You need 200 mL of 0.3 mol/L solution and have a 6 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 6 mol/Lc2 = 0.3 mol/LV2 = 200 mL
V1 = mL
4.Molarity & dilutionHard
200 mL of a 0.2 mol/L solution is mixed with 500 mL of a 1.2 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 75 g block of aluminium (c = 0.897 J/(g·°C)) at 110 °C is dropped into 400 g of water (c = 4.18 J/(g·°C)) at 15 °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 C4H10 (M = 58.12 g/mol) burns completely: 2C4H10 + 13O₂ → 8CO₂ + 10H₂O(l). Give q as a positive number.
0.4 mol/L of H₂ and 0.4 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 = 0.4 mol/LKc = 49
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.1 at the reaction temperature. A container starts with 0.2 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
[PCl₅]₀ = 0.2 mol/LKc = 0.1
[PCl5] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.05 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).
c = 0.05 mol/LKa = 6.3 × 10⁻⁵
pH =
12.pH & buffersHard
A buffer contains 0.25 mol/L hypochlorous acid and 0.05 mol/L of its sodium salt. Ka = 3 × 10⁻⁸. Find the pH pH of the buffer.
[HA] = 0.25[A⁻] = 0.05Ka = 3 × 10⁻⁸
pH =
13.Reaction kineticsMedium
A reaction with rate law −rA = k·CA² has k = 0.2 L/(mol·min) and starts at 0.5 mol/L. Find the concentration CA after 2 min.
k = 0.2 L/(mol·min)CA0 = 0.5 mol/Lt = 2 min
CA = mol/L
14.Reaction kineticsHard
Two experiments measure the initial rate of A → products. With CA0 = 0.2 mol/L the rate is 0.002 mol/(L·s); with CA0 = 0.4 mol/L it is 0.00283 mol/(L·s). Find the reaction order n with respect to A.
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(Na) = 20/22.99 = 0.8699 mol
n(Cl2) = 8/70.9 = 0.1128 mol
Limiting reactant
0.8699/2 = 0.435, 0.1128/1 = 0.1128 → Cl2
Mass of product
n(NaCl) = 0.1128·2 = 0.2257 mol
m = 0.2257·58.44 = 13.19 g
Answer
m = 13.19 g
2.Stoichiometry
Answer: Y = 69.99 %
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(Mg) = 30/24.31 = 1.234 mol
m = 1.234·2/2·40.3 = 49.73 g
Percent yield
Y = 34.81/49.73·100 = 69.99%
Answer
Y = 69.99%
3.Molarity & dilution
Answer: V1 = 10 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 = 0.3·200/6 = 10 mL
Answer
V1 = 10 mL
4.Molarity & dilution
Answer: c = 0.9143 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 = 0.2·200 + 1.2·500 = 640 mmol
Concentration of the mixture
c = 640/700 = 0.9143 mol/L
Answer
c = 0.9143 mol/L
5.Ideal gas law
Answer: V = 48.08 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 220/28.01 = 7.854 mol
T = 95 + 273.15 = 368.1 K
Volume
V = nRT/P = 7.854·8.314·368.1/500 = 48.08 L
Answer
V = 48.08 L
6.Ideal gas law
Answer: ρ = 7.944 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 160 + 273.15 = 433.1 K
Density
ρ = 650·44.01/(8.314·433.1) = 7.944 kg/m³
Answer
ρ = 7.944 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 18.67 °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
75·0.897·(110 − Tf) = 400·4.18·(Tf − 15)
Solve for the final temperature
Tf = (67.28·110 + 1672·15)/(67.28 + 1672) = 18.67 °C
Answer
Tf = 18.67 °C
8.Calorimetry & enthalpy
Answer: q = 1238 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
0.1 = x²/(0.2 − x) → x² + 0.1·x − 0.02 = 0
Solve the quadratic
x = (−0.1 + √(0.1² + 4·0.02))/2 = 0.1 mol/L
Equilibrium concentration
[PCl₅] = 0.2 − 0.1 = 0.1 mol/L
Answer
[PCl5] = 0.1 mol/L
11.pH & buffers
Answer: pH = 2.759
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 = 0.001744 mol/L
pH
pH = −log(0.001744) = 2.759
Answer
pH = 2.759
12.pH & buffers
Answer: pH = 6.824
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(0.05/0.25) = 6.824
Answer
pH = 6.824
13.Reaction kinetics
Answer: CA = 0.4167 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 = 2 + 0.2·2 = 2.4 L/mol
CA = 1/2.4 = 0.4167 mol/L
Answer
CA = 0.4167 mol/L
14.Reaction kinetics
Answer: n = 0.5008
Method of initial rates: for r = k·Cⁿ, comparing two runs that differ only in the starting concentration removes k: r₂/r₁ = (C₂/C₁)ⁿ. Doubling C doubles the rate for first order and quadruples it for second order.