10 g of N2 and 64 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 Fe2O3 + 3CO → 2Fe + 3CO2, 6 g of Fe2O3 reacts completely (the other reactants are in excess) and 4.51 g of CO2 is collected. Find the percent yield Y.
M(Fe2O3) = 159.69 g/molM(CO2) = 44.01 g/molactual: 4.51 g
Y = %
3.Molarity & dilutionMedium
You need 200 mL of 0.1 mol/L solution and have a 2 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 2 mol/Lc2 = 0.1 mol/LV2 = 200 mL
V1 = mL
4.Molarity & dilutionHard
300 mL of a 0.4 mol/L solution is mixed with 200 mL of a 0.6 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 75 g block of copper (c = 0.385 J/(g·°C)) at 80 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 19 °C in an insulated cup. Find the final temperature Tf.
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 [HI].
[H₂]₀ = [I₂]₀ = 0.1 mol/LKc = 9
[HI] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.04 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 = 0.04
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.01 mol/L solution of benzoic acid (Ka = 6.3 × 10⁻⁵).
c = 0.01 mol/LKa = 6.3 × 10⁻⁵
pH =
12.pH & buffersHard
A 1.00 L buffer contains 0.25 mol acetic acid and 0.25 mol of its conjugate base (Ka = 1.8 × 10⁻⁵). 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(N2) = 10/28.02 = 0.3569 mol
n(H2) = 64/2.016 = 31.75 mol
Limiting reactant
0.3569/1 = 0.3569, 31.75/3 = 10.58 → N2
Mass of product
n(NH3) = 0.3569·2 = 0.7138 mol
m = 0.7138·17.03 = 12.16 g
Answer
m = 12.16 g
2.Stoichiometry
Answer: Y = 90.91 %
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(Fe2O3) = 6/159.69 = 0.03757 mol
m = 0.03757·3/1·44.01 = 4.961 g
Percent yield
Y = 4.51/4.961·100 = 90.91%
Answer
Y = 90.91%
3.Molarity & dilution
Answer: V1 = 10 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 = 0.1·200/2 = 10 mL
Answer
V1 = 10 mL
4.Molarity & dilution
Answer: c = 0.48 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 = 0.4·300 + 0.6·200 = 240 mmol
Concentration of the mixture
c = 240/500 = 0.48 mol/L
Answer
c = 0.48 mol/L
5.Ideal gas law
Answer: P = 2744 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 140/4.003 = 34.97 mol
T = 10 + 273.15 = 283.1 K
Pressure
P = nRT/V = 34.97·8.314·283.1/0.03 = 2744 kPa
Answer
P = 2744 kPa
6.Ideal gas law
Answer: ρ = 1.56 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 20 + 273.15 = 293.1 K
Density
ρ = 950·4.003/(8.314·293.1) = 1.56 kg/m³
Answer
ρ = 1.56 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 20.64 °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
75·0.385·(80 − Tf) = 250·4.18·(Tf − 19)
Solve for the final temperature
Tf = (28.88·80 + 1045·19)/(28.88 + 1045) = 20.64 °C
Answer
Tf = 20.64 °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: [HI] = 0.12 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
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
[HI] = 2x = 0.12 mol/L
Answer
[HI] = 0.12 mol/L
10.Chemical equilibrium
Answer: [Cl2] = 0.07165 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/(0.2 − x) → x² + 0.04·x − 0.008 = 0
Solve the quadratic
x = (−0.04 + √(0.04² + 4·0.008))/2 = 0.07165 mol/L
Equilibrium concentration
[Cl₂] = x = 0.07165 mol/L
Answer
[Cl2] = 0.07165 mol/L
11.pH & buffers
Answer: pH = 3.118
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 = 0.0007629 mol/L
pH
pH = −log(0.0007629) = 3.118
Answer
pH = 3.118
12.pH & buffers
Answer: pH = 4.814
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) = 0.25 − 0.02 = 0.23 mol
n(A⁻) = 0.25 + 0.02 = 0.27 mol
Henderson–Hasselbalch
pH = 4.745 + log(0.27/0.23) = 4.814
Answer
pH = 4.814
13.Reaction kinetics
Answer: t = 6.931 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 = 0.25
Solve for the time
t = −ln(1 − X)/k = −ln(0.25)/0.2 = 6.931 min
Answer
t = 6.931 min
14.Reaction kinetics
Answer: k2 = 0.01044 min⁻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).