10 g of Mg and 40 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?
M(Mg) = 24.31M(O2) = 32M(MgO) = 40.3 g/mol
m = g
2.StoichiometryHard
In the reaction 2Na + Cl2 → 2NaCl, 8 g of Na reacts completely (the other reactants are in excess) and 18.51 g of NaCl is collected. Find the percent yield Y.
M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 18.51 g
Y = %
3.Molarity & dilutionMedium
You need 100 mL of 0.25 mol/L solution and have a 2 mol/L stock solution. How much water Vwater do you add to the stock solution?
c1 = 2 mol/Lc2 = 0.25 mol/LV2 = 100 mL
Vwater = mL
4.Molarity & dilutionHard
300 mL of a 1 mol/L solution is mixed with 250 mL of a 0.05 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
c1 = 1 mol/LV1 = 300 mLc2 = 0.05 mol/LV2 = 250 mL
c = mol/L
5.Ideal gas lawMedium
What volume V does 280 g of CO₂ occupy at 550 kPa and 35 °C? Treat it as an ideal gas.
A 250 g block of copper (c = 0.385 J/(g·°C)) at 150 °C is dropped into 250 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 25 g of C3H8 (M = 44.1 g/mol) burns completely: C3H8 + 5O₂ → 3CO₂ + 4H₂O(l). Give q as a positive number.
0.1 mol/L of H₂ and 0.1 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 64. Find the equilibrium concentration [HI].
[H₂]₀ = [I₂]₀ = 0.1 mol/LKc = 64
[HI] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.5 at the reaction temperature. A container starts with 0.5 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
[PCl₅]₀ = 0.5 mol/LKc = 0.5
[PCl5] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.25 mol/L solution of hydrofluoric acid (Ka = 0.00068).
c = 0.25 mol/LKa = 0.00068
pH =
12.pH & buffersHard
A 1.00 L buffer contains 0.1 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(Mg) = 10/24.31 = 0.4114 mol
n(O2) = 40/32 = 1.25 mol
Limiting reactant
0.4114/2 = 0.2057, 1.25/1 = 1.25 → Mg
Mass of product
n(MgO) = 0.2057·2 = 0.4114 mol
m = 0.4114·40.3 = 16.58 g
Answer
m = 16.58 g
2.Stoichiometry
Answer: Y = 91.02 %
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) = 8/22.99 = 0.348 mol
m = 0.348·2/2·58.44 = 20.34 g
Percent yield
Y = 18.51/20.34·100 = 91.02%
Answer
Y = 91.02%
3.Molarity & dilution
Answer: Vwater = 87.5 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 = 0.25·100/2 = 12.5 mL
Water to add
Vwater = 100 − 12.5 = 87.5 mL
Answer
Vwater = 87.5 mL
4.Molarity & dilution
Answer: c = 0.5682 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 = 1·300 + 0.05·250 = 312.5 mmol
Concentration of the mixture
c = 312.5/550 = 0.5682 mol/L
Answer
c = 0.5682 mol/L
5.Ideal gas law
Answer: V = 29.64 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 280/44.01 = 6.362 mol
T = 35 + 273.15 = 308.1 K
Volume
V = nRT/P = 6.362·8.314·308.1/550 = 29.64 L
Answer
V = 29.64 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 = 27.3 °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
250·0.385·(150 − Tf) = 250·4.18·(Tf − 16)
Solve for the final temperature
Tf = (96.25·150 + 1045·16)/(96.25 + 1045) = 27.3 °C
Answer
Tf = 27.3 °C
8.Calorimetry & enthalpy
Answer: q = 1258 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.
Equilibrium expression
64 = (2x)²/(0.1 − x)²
Take the square root
8 = 2x/(0.1 − x) → x = 0.1·8/(2 + 8) = 0.08 mol/L
Equilibrium concentration
[HI] = 2x = 0.16 mol/L
Answer
[HI] = 0.16 mol/L
10.Chemical equilibrium
Answer: [PCl5] = 0.191 mol/L
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/(0.5 − x) → x² + 0.5·x − 0.25 = 0
Solve the quadratic
x = (−0.5 + √(0.5² + 4·0.25))/2 = 0.309 mol/L
Equilibrium concentration
[PCl₅] = 0.5 − 0.309 = 0.191 mol/L
Answer
[PCl5] = 0.191 mol/L
11.pH & buffers
Answer: pH = 1.896
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 = 0.0127 mol/L
pH
pH = −log(0.0127) = 1.896
Answer
pH = 1.896
12.pH & buffers
Answer: pH = 4.678
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) = 0.1 − 0.05 = 0.05 mol
n(A⁻) = 0.1 + 0.05 = 0.15 mol
Henderson–Hasselbalch
pH = 4.201 + log(0.15/0.05) = 4.678
Answer
pH = 4.678
13.Reaction kinetics
Answer: t = 23.03 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 = 0.1
Solve for the time
t = −ln(1 − X)/k = −ln(0.1)/0.1 = 23.03 min
Answer
t = 23.03 min
14.Reaction kinetics
Answer: k2 = 0.08508 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).