30 g of C3H8 and 16 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of H2O can form?
M(C3H8) = 44.1M(O2) = 32M(H2O) = 18.02 g/mol
m = g
2.StoichiometryHard
In the reaction 2Al + 3Cl2 → 2AlCl3, 12 g of Al reacts completely (the other reactants are in excess) and 52.19 g of AlCl3 is collected. Find the percent yield Y.
M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 52.19 g
Y = %
3.Molarity & dilutionMedium
You need 500 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 = 500 mL
Vwater = mL
4.Molarity & dilutionHard
200 mL of a 1 mol/L solution is mixed with 500 mL of a 0.3 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
c1 = 1 mol/LV1 = 200 mLc2 = 0.3 mol/LV2 = 500 mL
c = mol/L
5.Ideal gas lawMedium
What volume V does 60 g of CO₂ occupy at 700 kPa and 85 °C? Treat it as an ideal gas.
A 50 g block of iron (c = 0.449 J/(g·°C)) at 160 °C is dropped into 150 g of water (c = 4.18 J/(g·°C)) at 16 °C in an insulated cup. Find the final temperature Tf.
1 mol/L of H₂ and 1 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 16. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 1 mol/LKc = 16
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.1 at the reaction temperature. A container starts with 1.5 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 1.5 mol/LKc = 0.1
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.5 mol/L solution of formic acid (Ka = 0.00018).
c = 0.5 mol/LKa = 0.00018
pH =
12.pH & buffersHard
A buffer contains 0.5 mol/L hypochlorous acid and 0.2 mol/L of its sodium salt. Ka = 3 × 10⁻⁸. Find the pH pH of the buffer.
[HA] = 0.5[A⁻] = 0.2Ka = 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 1 mol/L. Find the concentration CA after 10 min.
k = 0.2 L/(mol·min)CA0 = 1 mol/Lt = 10 min
CA = mol/L
14.Reaction kineticsHard
A first-order reaction has a rate constant of 0.05 s⁻¹ at 30 °C and an activation energy of 75 kJ/mol. Find the pre-exponential factor A in the Arrhenius equation.
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(C3H8) = 30/44.1 = 0.6803 mol
n(O2) = 16/32 = 0.5 mol
Limiting reactant
0.6803/1 = 0.6803, 0.5/5 = 0.1 → O2
Mass of product
n(H2O) = 0.1·4 = 0.4 mol
m = 0.4·18.02 = 7.208 g
Answer
m = 7.208 g
2.Stoichiometry
Answer: Y = 88.01 %
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(Al) = 12/26.98 = 0.4448 mol
m = 0.4448·2/2·133.33 = 59.3 g
Percent yield
Y = 52.19/59.3·100 = 88.01%
Answer
Y = 88.01%
3.Molarity & dilution
Answer: Vwater = 437.5 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 = 0.25·500/2 = 62.5 mL
Water to add
Vwater = 500 − 62.5 = 437.5 mL
Answer
Vwater = 437.5 mL
4.Molarity & dilution
Answer: c = 0.5 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 = 1·200 + 0.3·500 = 350 mmol
Concentration of the mixture
c = 350/700 = 0.5 mol/L
Answer
c = 0.5 mol/L
5.Ideal gas law
Answer: V = 5.799 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 60/44.01 = 1.363 mol
T = 85 + 273.15 = 358.1 K
Volume
V = nRT/P = 1.363·8.314·358.1/700 = 5.799 L
Answer
V = 5.799 L
6.Ideal gas law
Answer: V2 = 0.9428 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Absolute temperatures
T1 = 308.1 K, T2 = 363.1 K
Combined gas law
V2 = V1·(P1/P2)·(T2/T1) = 2·(300/750)·(363.1/308.1) = 0.9428 L
Answer
V2 = 0.9428 L
7.Calorimetry & enthalpy
Answer: Tf = 20.98 °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
50·0.449·(160 − Tf) = 150·4.18·(Tf − 16)
Solve for the final temperature
Tf = (22.45·160 + 627·16)/(22.45 + 627) = 20.98 °C
Answer
Tf = 20.98 °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] = 0.3333 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
16 = (2x)²/(1 − x)²
Take the square root
4 = 2x/(1 − x) → x = 1·4/(2 + 4) = 0.6667 mol/L
Equilibrium concentration
[H₂] = 1 − x = 0.3333 mol/L
Answer
[H2] = 0.3333 mol/L
10.Chemical equilibrium
Answer: [Cl2] = 0.3405 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
0.1 = x²/(1.5 − x) → x² + 0.1·x − 0.15 = 0
Solve the quadratic
x = (−0.1 + √(0.1² + 4·0.15))/2 = 0.3405 mol/L
Equilibrium concentration
[Cl₂] = x = 0.3405 mol/L
Answer
[Cl2] = 0.3405 mol/L
11.pH & buffers
Answer: pH = 2.027
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 = 0.009397 mol/L
pH
pH = −log(0.009397) = 2.027
Answer
pH = 2.027
12.pH & buffers
Answer: pH = 7.125
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(0.2/0.5) = 7.125
Answer
pH = 7.125
13.Reaction kinetics
Answer: CA = 0.3333 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 = 1 + 0.2·10 = 3 L/mol
CA = 1/3 = 0.3333 mol/L
Answer
CA = 0.3333 mol/L
14.Reaction kinetics
Answer: A = 4.192 × 10¹¹ s⁻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).