8 g of Fe2O3 and 40 g of CO react according to Fe2O3 + 3CO → 2Fe + 3CO2. What mass m of Fe can form?
M(Fe2O3) = 159.69M(CO) = 28.01M(Fe) = 55.85 g/mol
m = g
2.StoichiometryHard
In the reaction 2Mg + O2 → 2MgO, 24 g of Mg reacts completely (the other reactants are in excess) and 37.4 g of MgO is collected. Find the percent yield Y.
M(Mg) = 24.31 g/molM(MgO) = 40.3 g/molactual: 37.4 g
Y = %
3.Molarity & dilutionMedium
You need 250 mL of 0.25 mol/L solution and have a 1 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 1 mol/Lc2 = 0.25 mol/LV2 = 250 mL
V1 = mL
4.Molarity & dilutionHard
150 mL of a 1 mol/L solution is mixed with 200 mL of a 3 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
c1 = 1 mol/LV1 = 150 mLc2 = 3 mol/LV2 = 200 mL
c = mol/L
5.Ideal gas lawMedium
What volume V does 120 g of O₂ occupy at 500 kPa and 20 °C? Treat it as an ideal gas.
A 50 g block of zinc (c = 0.388 J/(g·°C)) at 120 °C is dropped into 100 g of water (c = 4.18 J/(g·°C)) at 18 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find the heat q released when 50 g of C2H6 (M = 30.07 g/mol) burns completely: 2C2H6 + 7O₂ → 4CO₂ + 6H₂O(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 = 9. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.1 mol/LKc = 9
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 at the reaction temperature. A container starts with 0.2 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 0.2 mol/LKc = 0.08
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.01 mol/L solution of hydrofluoric acid (Ka = 0.00068).
c = 0.01 mol/LKa = 0.00068
pH =
12.pH & buffersHard
A 1.00 L buffer contains 0.3 mol formic acid and 0.2 mol of its conjugate base (Ka = 0.00018). 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(Fe2O3) = 8/159.69 = 0.0501 mol
n(CO) = 40/28.01 = 1.428 mol
Limiting reactant
0.0501/1 = 0.0501, 1.428/3 = 0.476 → Fe2O3
Mass of product
n(Fe) = 0.0501·2 = 0.1002 mol
m = 0.1002·55.85 = 5.596 g
Answer
m = 5.596 g
2.Stoichiometry
Answer: Y = 94 %
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) = 24/24.31 = 0.9872 mol
m = 0.9872·2/2·40.3 = 39.79 g
Percent yield
Y = 37.4/39.79·100 = 94%
Answer
Y = 94%
3.Molarity & dilution
Answer: V1 = 62.5 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 = 0.25·250/1 = 62.5 mL
Answer
V1 = 62.5 mL
4.Molarity & dilution
Answer: c = 2.143 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 = 1·150 + 3·200 = 750 mmol
Concentration of the mixture
c = 750/350 = 2.143 mol/L
Answer
c = 2.143 mol/L
5.Ideal gas law
Answer: V = 18.28 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 120/32 = 3.75 mol
T = 20 + 273.15 = 293.1 K
Volume
V = nRT/P = 3.75·8.314·293.1/500 = 18.28 L
Answer
V = 18.28 L
6.Ideal gas law
Answer: V2 = 6.142 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Absolute temperatures
T1 = 313.1 K, T2 = 423.1 K
Combined gas law
V2 = V1·(P1/P2)·(T2/T1) = 10·(250/550)·(423.1/313.1) = 6.142 L
Answer
V2 = 6.142 L
7.Calorimetry & enthalpy
Answer: Tf = 22.52 °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
50·0.388·(120 − Tf) = 100·4.18·(Tf − 18)
Solve for the final temperature
Tf = (19.4·120 + 418·18)/(19.4 + 418) = 22.52 °C
Answer
Tf = 22.52 °C
8.Calorimetry & enthalpy
Answer: q = 2593 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
9 = (2x)²/(0.1 − x)²
Take the square root
3 = 2x/(0.1 − x) → x = 0.1·3/(2 + 3) = 0.06 mol/L
Equilibrium concentration
[H₂] = 0.1 − x = 0.04 mol/L
Answer
[H2] = 0.04 mol/L
10.Chemical equilibrium
Answer: [Cl2] = 0.09266 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
0.08 = x²/(0.2 − x) → x² + 0.08·x − 0.016 = 0
Solve the quadratic
x = (−0.08 + √(0.08² + 4·0.016))/2 = 0.09266 mol/L
Equilibrium concentration
[Cl₂] = x = 0.09266 mol/L
Answer
[Cl2] = 0.09266 mol/L
11.pH & buffers
Answer: pH = 2.64
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/(0.01 − x) → x = 0.00229 mol/L
pH
pH = −log(0.00229) = 2.64
Answer
pH = 2.64
12.pH & buffers
Answer: pH = 3.64
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 − 0.02 = 0.28 mol
n(A⁻) = 0.2 + 0.02 = 0.22 mol
Henderson–Hasselbalch
pH = 3.745 + log(0.22/0.28) = 3.64
Answer
pH = 3.64
13.Reaction kinetics
Answer: t = 32.19 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 = 0.2
Solve for the time
t = −ln(1 − X)/k = −ln(0.2)/0.05 = 32.19 min
Answer
t = 32.19 min
14.Reaction kinetics
Answer: Ea = 69.94 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 = 303.1 K, T2 = 318.1 K
Solve Arrhenius for Ea
Ea = −R·ln(k2/k1)/(1/T2 − 1/T1) = −8.314·1.308/(-1.555e-4) = 69940 J/mol