6 g of N2 and 80 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 2Al + 3Cl2 → 2AlCl3, 8 g of Al reacts completely (the other reactants are in excess) and 27.67 g of AlCl3 is collected. Find the percent yield Y.
M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 27.67 g
Y = %
3.Molarity & dilutionMedium
You need 200 mL of 0.3 mol/L solution and have a 0.5 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 0.5 mol/Lc2 = 0.3 mol/LV2 = 200 mL
V1 = mL
4.Molarity & dilutionHard
250 mL of a 0.1 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 200 g block of iron (c = 0.449 J/(g·°C)) at 200 °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.
Use the standard enthalpies of formation to find the heat q released when 50 g of C2H5OH (M = 46.07 g/mol) burns completely: C2H5OH + 3O₂ → 2CO₂ + 3H₂O(l). Give q as a positive number.
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 = 0.08 at the reaction temperature. A container starts with 0.4 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 0.4 mol/LKc = 0.08
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.2 mol/L solution of lactic acid (Ka = 0.00014).
c = 0.2 mol/LKa = 0.00014
pH =
12.pH & buffersHard
A buffer contains 0.1 mol/L benzoic acid and 0.1 mol/L of its sodium salt. Ka = 6.3 × 10⁻⁵. Find the pH pH of the buffer.
[HA] = 0.1[A⁻] = 0.1Ka = 6.3 × 10⁻⁵
pH =
13.Reaction kineticsMedium
A first-order reaction in a batch reactor has k = 0.02 min⁻¹. How long does it take to reach 99% conversion? Find t.
k = 0.02 min⁻¹X = 99 %
t = min
14.Reaction kineticsHard
A reaction has a rate constant of 0.002 min⁻¹ at 35 °C and an activation energy of 40 kJ/mol. Find the rate constant k2 at 55 °C.
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) = 6/28.02 = 0.2141 mol
n(H2) = 80/2.016 = 39.68 mol
Limiting reactant
0.2141/1 = 0.2141, 39.68/3 = 13.23 → N2
Mass of product
n(NH3) = 0.2141·2 = 0.4283 mol
m = 0.4283·17.03 = 7.293 g
Answer
m = 7.293 g
2.Stoichiometry
Answer: Y = 69.99 %
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 = 27.67/39.53·100 = 69.99%
Answer
Y = 69.99%
3.Molarity & dilution
Answer: V1 = 120 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 = 0.3·200/0.5 = 120 mL
Answer
V1 = 120 mL
4.Molarity & dilution
Answer: c = 0.3222 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 = 0.1·250 + 0.6·200 = 145 mmol
Concentration of the mixture
c = 145/450 = 0.3222 mol/L
Answer
c = 0.3222 mol/L
5.Ideal gas law
Answer: V = 18.49 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 170/28.01 = 6.069 mol
T = 20 + 273.15 = 293.1 K
Volume
V = nRT/P = 6.069·8.314·293.1/800 = 18.49 L
Answer
V = 18.49 L
6.Ideal gas law
Answer: V2 = 3.424 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Absolute temperatures
T1 = 313.1 K, T2 = 343.1 K
Combined gas law
V2 = V1·(P1/P2)·(T2/T1) = 5·(250/400)·(343.1/313.1) = 3.424 L
Answer
V2 = 3.424 L
7.Calorimetry & enthalpy
Answer: Tf = 33.32 °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
200·0.449·(200 − Tf) = 250·4.18·(Tf − 19)
Solve for the final temperature
Tf = (89.8·200 + 1045·19)/(89.8 + 1045) = 33.32 °C
Answer
Tf = 33.32 °C
8.Calorimetry & enthalpy
Answer: q = 1483 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
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] = 0.1433 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/(0.4 − x) → x² + 0.08·x − 0.032 = 0
Solve the quadratic
x = (−0.08 + √(0.08² + 4·0.032))/2 = 0.1433 mol/L
Equilibrium concentration
[Cl₂] = x = 0.1433 mol/L
Answer
[Cl2] = 0.1433 mol/L
11.pH & buffers
Answer: pH = 2.282
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 = 0.005222 mol/L
pH
pH = −log(0.005222) = 2.282
Answer
pH = 2.282
12.pH & buffers
Answer: pH = 4.201
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 = 4.201
Henderson–Hasselbalch
pH = 4.201 + log(0.1/0.1) = 4.201
Answer
pH = 4.201
13.Reaction kinetics
Answer: t = 230.3 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 = 0.01
Solve for the time
t = −ln(1 − X)/k = −ln(0.01)/0.02 = 230.3 min
Answer
t = 230.3 min
14.Reaction kinetics
Answer: k2 = 0.00518 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).