20 g of H2 and 32 g of O2 react according to 2H2 + O2 → 2H2O. What mass m of H2O can form?
M(H2) = 2.016M(O2) = 32M(H2O) = 18.02 g/mol
m = g
2.StoichiometryHard
In the reaction 2Al + 3Cl2 → 2AlCl3, 8 g of Al reacts completely (the other reactants are in excess) and 35.98 g of AlCl3 is collected. Find the percent yield Y.
M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 35.98 g
Y = %
3.Molarity & dilutionMedium
You need 1000 mL of 0.4 mol/L solution and have a 6 mol/L stock solution. How much water Vwater do you add to the stock solution?
c1 = 6 mol/Lc2 = 0.4 mol/LV2 = 1000 mL
Vwater = mL
4.Molarity & dilutionHard
200 mL of a 0.4 mol/L solution is mixed with 400 mL of a 1.2 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 200 g block of copper (c = 0.385 J/(g·°C)) at 130 °C is dropped into 300 g of water (c = 4.18 J/(g·°C)) at 18 °C in an insulated cup. Find the final temperature Tf.
0.25 mol/L of H₂ and 0.25 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 = 0.25 mol/LKc = 16
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.02 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 = 0.02
[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 buffer contains 0.2 mol/L lactic acid and 0.2 mol/L of its sodium salt. Ka = 0.00014. Find the pH pH of the buffer.
[HA] = 0.2[A⁻] = 0.2Ka = 0.00014
pH =
13.Reaction kineticsMedium
A zero-order reaction has k = 0.005 mol/(L·min) and starts at 1 mol/L. How long does it take to reach 20% conversion? Find t.
k = 0.005 mol/(L·min)CA0 = 1 mol/LX = 20 %
t = min
14.Reaction kineticsHard
A first-order reaction has a rate constant of 0.005 s⁻¹ at 25 °C and an activation energy of 100 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(H2) = 20/2.016 = 9.921 mol
n(O2) = 32/32 = 1 mol
Limiting reactant
9.921/2 = 4.96, 1/1 = 1 → O2
Mass of product
n(H2O) = 1·2 = 2 mol
m = 2·18.02 = 36.04 g
Answer
m = 36.04 g
2.Stoichiometry
Answer: Y = 91.01 %
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) = 8/26.98 = 0.2965 mol
m = 0.2965·2/2·133.33 = 39.53 g
Percent yield
Y = 35.98/39.53·100 = 91.01%
Answer
Y = 91.01%
3.Molarity & dilution
Answer: Vwater = 933.3 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 = 0.4·1000/6 = 66.67 mL
Water to add
Vwater = 1000 − 66.67 = 933.3 mL
Answer
Vwater = 933.3 mL
4.Molarity & dilution
Answer: c = 0.9333 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 = 0.4·200 + 1.2·400 = 560 mmol
Concentration of the mixture
c = 560/600 = 0.9333 mol/L
Answer
c = 0.9333 mol/L
5.Ideal gas law
Answer: P = 387.7 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 140/44.01 = 3.181 mol
T = 20 + 273.15 = 293.1 K
Pressure
P = nRT/V = 3.181·8.314·293.1/0.02 = 387.7 kPa
Answer
P = 387.7 kPa
6.Ideal gas law
Answer: ρ = 15.27 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 10 + 273.15 = 283.1 K
Density
ρ = 900·39.95/(8.314·283.1) = 15.27 kg/m³
Answer
ρ = 15.27 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 24.48 °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
200·0.385·(130 − Tf) = 300·4.18·(Tf − 18)
Solve for the final temperature
Tf = (77·130 + 1254·18)/(77 + 1254) = 24.48 °C
Answer
Tf = 24.48 °C
8.Calorimetry & enthalpy
Answer: ΔH° = −3119 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] = 0.08333 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
0.02 = x²/(0.2 − x) → x² + 0.02·x − 0.004 = 0
Solve the quadratic
x = (−0.02 + √(0.02² + 4·0.004))/2 = 0.05403 mol/L
Equilibrium concentration
[PCl₅] = 0.2 − 0.05403 = 0.146 mol/L
Answer
[PCl5] = 0.146 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 = 3.854
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 = 3.854
Henderson–Hasselbalch
pH = 3.854 + log(0.2/0.2) = 3.854
Answer
pH = 3.854
13.Reaction kinetics
Answer: t = 40 min
Integrated rate laws: zero order: C = C₀ − kt (straight line, t½ = C₀/(2k)); first order: ln(C₀/C) = kt (t½ = ln 2/k); second order: 1/C − 1/C₀ = kt (t½ = 1/(kC₀)). Conversion X means C = C₀(1 − X).
Amount converted
CA0·X = 1·0.2 = 0.2 mol/L
Time
t = CA0·X/k = 0.2/0.005 = 40 min
Answer
t = 40 min
14.Reaction kinetics
Answer: A = 1.656 × 10¹⁵ 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).