30 g of N2 and 8 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 CaCO3 → CaO + CO2, 20 g of CaCO3 reacts completely (the other reactants are in excess) and 9.19 g of CaO is collected. Find the percent yield Y.
M(CaCO3) = 100.09 g/molM(CaO) = 56.08 g/molactual: 9.19 g
Y = %
3.Molarity & dilutionMedium
You need 100 mL of 0.5 mol/L solution and have a 6 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 6 mol/Lc2 = 0.5 mol/LV2 = 100 mL
V1 = mL
4.Molarity & dilutionHard
300 mL of a 1.5 mol/L solution is mixed with 100 mL of a 0.3 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 200 g block of lead (c = 0.129 J/(g·°C)) at 130 °C is dropped into 250 g of water (c = 4.18 J/(g·°C)) at 25 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find the heat q released when 100 g of CH4 (M = 16.04 g/mol) burns completely: CH4 + 2O₂ → CO₂ + 2H₂O(l). Give q as a positive number.
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 = 9. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.25 mol/LKc = 9
[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 lactic acid (Ka = 0.00014).
c = 0.5 mol/LKa = 0.00014
pH =
12.pH & buffersHard
A buffer contains 0.2 mol/L benzoic acid and 0.3 mol/L of its sodium salt. Ka = 6.3 × 10⁻⁵. Find the pH pH of the buffer.
[HA] = 0.2[A⁻] = 0.3Ka = 6.3 × 10⁻⁵
pH =
13.Reaction kineticsMedium
A first-order reaction in a batch reactor has k = 0.08 min⁻¹. How long does it take to reach 90% conversion? Find t.
k = 0.08 min⁻¹X = 90 %
t = min
14.Reaction kineticsHard
A reaction has a rate constant of 0.01 min⁻¹ at 30 °C and an activation energy of 50 kJ/mol. Find the rate constant k2 at 40 °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) = 30/28.02 = 1.071 mol
n(H2) = 8/2.016 = 3.968 mol
Limiting reactant
1.071/1 = 1.071, 3.968/3 = 1.323 → N2
Mass of product
n(NH3) = 1.071·2 = 2.141 mol
m = 2.141·17.03 = 36.47 g
Answer
m = 36.47 g
2.Stoichiometry
Answer: Y = 82.01 %
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(CaCO3) = 20/100.09 = 0.1998 mol
m = 0.1998·1/1·56.08 = 11.21 g
Percent yield
Y = 9.19/11.21·100 = 82.01%
Answer
Y = 82.01%
3.Molarity & dilution
Answer: V1 = 8.333 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 = 0.5·100/6 = 8.333 mL
Answer
V1 = 8.333 mL
4.Molarity & dilution
Answer: c = 1.2 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 = 1.5·300 + 0.3·100 = 480 mmol
Concentration of the mixture
c = 480/400 = 1.2 mol/L
Answer
c = 1.2 mol/L
5.Ideal gas law
Answer: P = 1916 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 270/32 = 8.438 mol
T = 0 + 273.15 = 273.1 K
Pressure
P = nRT/V = 8.438·8.314·273.1/0.01 = 1916 kPa
Answer
P = 1916 kPa
6.Ideal gas law
Answer: V2 = 4.949 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Absolute temperatures
T1 = 298.1 K, T2 = 413.1 K
Combined gas law
V2 = V1·(P1/P2)·(T2/T1) = 10·(250/700)·(413.1/298.1) = 4.949 L
Answer
V2 = 4.949 L
7.Calorimetry & enthalpy
Answer: Tf = 27.53 °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
200·0.129·(130 − Tf) = 250·4.18·(Tf − 25)
Solve for the final temperature
Tf = (25.8·130 + 1045·25)/(25.8 + 1045) = 27.53 °C
Answer
Tf = 27.53 °C
8.Calorimetry & enthalpy
Answer: q = 5550 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.
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 = 2.081
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 = 0.008297 mol/L
pH
pH = −log(0.008297) = 2.081
Answer
pH = 2.081
12.pH & buffers
Answer: pH = 4.377
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(0.3/0.2) = 4.377
Answer
pH = 4.377
13.Reaction kinetics
Answer: t = 28.78 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 = 0.1
Solve for the time
t = −ln(1 − X)/k = −ln(0.1)/0.08 = 28.78 min
Answer
t = 28.78 min
14.Reaction kinetics
Answer: k2 = 0.01884 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).