6 g of Mg and 16 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, 6 g of Na reacts completely (the other reactants are in excess) and 8.39 g of NaCl is collected. Find the percent yield Y.
M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 8.39 g
Y = %
3.Molarity & dilutionMedium
You need 100 mL of 0.4 mol/L solution and have a 2.5 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 2.5 mol/Lc2 = 0.4 mol/LV2 = 100 mL
V1 = mL
4.Molarity & dilutionHard
200 mL of a 1.5 mol/L solution is mixed with 50 mL of a 0.6 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
c1 = 1.5 mol/LV1 = 200 mLc2 = 0.6 mol/LV2 = 50 mL
c = mol/L
5.Ideal gas lawMedium
What volume V does 490 g of CH₄ occupy at 700 kPa and 45 °C? Treat it as an ideal gas.
A 75 g block of zinc (c = 0.388 J/(g·°C)) at 110 °C is dropped into 300 g of water (c = 4.18 J/(g·°C)) at 23 °C in an insulated cup. Find the final temperature Tf.
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 = 50. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.4 mol/LKc = 50
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.04 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 = 0.04
[PCl5] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.2 mol/L solution of formic acid (Ka = 0.00018).
c = 0.2 mol/LKa = 0.00018
pH =
12.pH & buffersHard
A 1.00 L buffer contains 0.2 mol hydrofluoric acid and 0.25 mol of its conjugate base (Ka = 0.00068). 0.02 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) = 6/24.31 = 0.2468 mol
n(O2) = 16/32 = 0.5 mol
Limiting reactant
0.2468/2 = 0.1234, 0.5/1 = 0.5 → Mg
Mass of product
n(MgO) = 0.1234·2 = 0.2468 mol
m = 0.2468·40.3 = 9.947 g
Answer
m = 9.947 g
2.Stoichiometry
Answer: Y = 55.01 %
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) = 6/22.99 = 0.261 mol
m = 0.261·2/2·58.44 = 15.25 g
Percent yield
Y = 8.39/15.25·100 = 55.01%
Answer
Y = 55.01%
3.Molarity & dilution
Answer: V1 = 16 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 = 0.4·100/2.5 = 16 mL
Answer
V1 = 16 mL
4.Molarity & dilution
Answer: c = 1.32 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 = 1.5·200 + 0.6·50 = 330 mmol
Concentration of the mixture
c = 330/250 = 1.32 mol/L
Answer
c = 1.32 mol/L
5.Ideal gas law
Answer: V = 115.4 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 490/16.04 = 30.55 mol
T = 45 + 273.15 = 318.1 K
Volume
V = nRT/P = 30.55·8.314·318.1/700 = 115.4 L
Answer
V = 115.4 L
6.Ideal gas law
Answer: ρ = 10.47 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 140 + 273.15 = 413.1 K
Density
ρ = 900·39.95/(8.314·413.1) = 10.47 kg/m³
Answer
ρ = 10.47 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 24.97 °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
75·0.388·(110 − Tf) = 300·4.18·(Tf − 23)
Solve for the final temperature
Tf = (29.1·110 + 1254·23)/(29.1 + 1254) = 24.97 °C
Answer
Tf = 24.97 °C
8.Calorimetry & enthalpy
Answer: ΔH° = −1367 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.
Negative: the reaction gives off heat (exothermic).
9.Chemical equilibrium
Answer: [H2] = 0.08819 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.
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
0.04 = x²/(0.5 − x) → x² + 0.04·x − 0.02 = 0
Solve the quadratic
x = (−0.04 + √(0.04² + 4·0.02))/2 = 0.1228 mol/L
Equilibrium concentration
[PCl₅] = 0.5 − 0.1228 = 0.3772 mol/L
Answer
[PCl5] = 0.3772 mol/L
11.pH & buffers
Answer: pH = 2.228
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/(0.2 − x) → x = 0.005911 mol/L
pH
pH = −log(0.005911) = 2.228
Answer
pH = 2.228
12.pH & buffers
Answer: pH = 3.344
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) = 0.2 − 0.02 = 0.18 mol
n(A⁻) = 0.25 + 0.02 = 0.27 mol
Henderson–Hasselbalch
pH = 3.167 + log(0.27/0.18) = 3.344
Answer
pH = 3.344
13.Reaction kinetics
Answer: t = 46.05 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)/0.05 = 46.05 min
Answer
t = 46.05 min
14.Reaction kinetics
Answer: Ea = 60 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 = 293.1 K, T2 = 323.1 K
Solve Arrhenius for Ea
Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·2.285/(-3.167e-4) = 60000 J/mol