16 g of N2 and 32 g of H2 react according to N2 + 3H2 → 2NH3. What mass m of NH3 can form?
M(N2) = 28.02M(H2) = 2.016M(NH3) = 17.03 g/mol
m = g
2.StoichiometryHard
In the reaction N2 + 3H2 → 2NH3, 8 g of N2 reacts completely (the other reactants are in excess) and 7.29 g of NH3 is collected. Find the percent yield Y.
M(N2) = 28.02 g/molM(NH3) = 17.03 g/molactual: 7.29 g
Y = %
3.Molarity & dilutionMedium
You need 100 mL of 0.2 mol/L solution and have a 2.5 mol/L stock solution. How much water Vwater do you add to the stock solution?
c1 = 2.5 mol/Lc2 = 0.2 mol/LV2 = 100 mL
Vwater = mL
4.Molarity & dilutionHard
50 mL of a 1 mol/L solution is mixed with 250 mL of a 1.2 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
c1 = 1 mol/LV1 = 50 mLc2 = 1.2 mol/LV2 = 250 mL
c = mol/L
5.Ideal gas lawMedium
What volume V does 470 g of CO₂ occupy at 500 kPa and 70 °C? Treat it as an ideal gas.
A 50 g block of zinc (c = 0.388 J/(g·°C)) at 140 °C is dropped into 200 g of water (c = 4.18 J/(g·°C)) at 22 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find the heat q released when 10 g of C3H8 (M = 44.1 g/mol) burns completely: C3H8 + 5O₂ → 3CO₂ + 4H₂O(l). Give q as a positive number.
0.2 mol/L of H₂ and 0.2 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 4. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.2 mol/LKc = 4
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.04 at the reaction temperature. A container starts with 1 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 1 mol/LKc = 0.04
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.25 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).
c = 0.25 mol/LKa = 6.3 × 10⁻⁵
pH =
12.pH & buffersHard
A buffer contains 0.2 mol/L acetic acid and 0.5 mol/L of its sodium salt. Ka = 1.8 × 10⁻⁵. Find the pH pH of the buffer.
[HA] = 0.2[A⁻] = 0.5Ka = 1.8 × 10⁻⁵
pH =
13.Reaction kineticsMedium
A reaction with rate law −rA = k·CA² has k = 0.4 L/(mol·min) and starts at 1.5 mol/L. Find the concentration CA after 30 min.
k = 0.4 L/(mol·min)CA0 = 1.5 mol/Lt = 30 min
CA = mol/L
14.Reaction kineticsHard
A first-order reaction has a rate constant of 0.1 s⁻¹ at 50 °C and an activation energy of 90 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(N2) = 16/28.02 = 0.571 mol
n(H2) = 32/2.016 = 15.87 mol
Limiting reactant
0.571/1 = 0.571, 15.87/3 = 5.291 → N2
Mass of product
n(NH3) = 0.571·2 = 1.142 mol
m = 1.142·17.03 = 19.45 g
Answer
m = 19.45 g
2.Stoichiometry
Answer: Y = 74.97 %
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(N2) = 8/28.02 = 0.2855 mol
m = 0.2855·2/1·17.03 = 9.724 g
Percent yield
Y = 7.29/9.724·100 = 74.97%
Answer
Y = 74.97%
3.Molarity & dilution
Answer: Vwater = 92 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 = 0.2·100/2.5 = 8 mL
Water to add
Vwater = 100 − 8 = 92 mL
Answer
Vwater = 92 mL
4.Molarity & dilution
Answer: c = 1.167 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 = 1·50 + 1.2·250 = 350 mmol
Concentration of the mixture
c = 350/300 = 1.167 mol/L
Answer
c = 1.167 mol/L
5.Ideal gas law
Answer: V = 60.94 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 470/44.01 = 10.68 mol
T = 70 + 273.15 = 343.1 K
Volume
V = nRT/P = 10.68·8.314·343.1/500 = 60.94 L
Answer
V = 60.94 L
6.Ideal gas law
Answer: ρ = 1.427 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 200 + 273.15 = 473.1 K
Density
ρ = 350·16.04/(8.314·473.1) = 1.427 kg/m³
Answer
ρ = 1.427 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 24.68 °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·(140 − Tf) = 200·4.18·(Tf − 22)
Solve for the final temperature
Tf = (19.4·140 + 836·22)/(19.4 + 836) = 24.68 °C
Answer
Tf = 24.68 °C
8.Calorimetry & enthalpy
Answer: q = 503.4 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
4 = (2x)²/(0.2 − x)²
Take the square root
2 = 2x/(0.2 − x) → x = 0.2·2/(2 + 2) = 0.1 mol/L
Equilibrium concentration
[H₂] = 0.2 − x = 0.1 mol/L
Answer
[H2] = 0.1 mol/L
10.Chemical equilibrium
Answer: [Cl2] = 0.181 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
0.04 = x²/(1 − x) → x² + 0.04·x − 0.04 = 0
Solve the quadratic
x = (−0.04 + √(0.04² + 4·0.04))/2 = 0.181 mol/L
Equilibrium concentration
[Cl₂] = x = 0.181 mol/L
Answer
[Cl2] = 0.181 mol/L
11.pH & buffers
Answer: pH = 2.405
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 = 0.003937 mol/L
pH
pH = −log(0.003937) = 2.405
Answer
pH = 2.405
12.pH & buffers
Answer: pH = 5.143
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 = 4.745
Henderson–Hasselbalch
pH = 4.745 + log(0.5/0.2) = 5.143
Answer
pH = 5.143
13.Reaction kinetics
Answer: CA = 0.07895 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 = 0.6667 + 0.4·30 = 12.67 L/mol
CA = 1/12.67 = 0.07895 mol/L
Answer
CA = 0.07895 mol/L
14.Reaction kinetics
Answer: A = 3.534 × 10¹³ 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).