24 g of CH4 and 20 g of O2 react according to CH4 + 2O2 → CO2 + 2H2O. What mass m of H2O can form?
M(CH4) = 16.04M(O2) = 32M(H2O) = 18.02 g/mol
m = g
2.StoichiometryHard
In the reaction CaCO3 → CaO + CO2, 10 g of CaCO3 reacts completely (the other reactants are in excess) and 4.37 g of CaO is collected. Find the percent yield Y.
M(CaCO3) = 100.09 g/molM(CaO) = 56.08 g/molactual: 4.37 g
Y = %
3.Molarity & dilutionMedium
You need 500 mL of 0.4 mol/L solution and have a 0.5 mol/L stock solution. How much water Vwater do you add to the stock solution?
c1 = 0.5 mol/Lc2 = 0.4 mol/LV2 = 500 mL
Vwater = mL
4.Molarity & dilutionHard
300 mL of a 0.2 mol/L solution is mixed with 400 mL of a 2.5 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 50 g block of zinc (c = 0.388 J/(g·°C)) at 160 °C is dropped into 300 g of water (c = 4.18 J/(g·°C)) at 24 °C in an insulated cup. Find the final temperature Tf.
Use the standard enthalpies of formation to find the heat q released when 25 g of C6H12O6 (M = 180.16 g/mol) burns completely: C6H12O6 + 6O₂ → 6CO₂ + 6H₂O(l). Give q as a positive number.
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 = 64. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.1 mol/LKc = 64
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.08 at the reaction temperature. A container starts with 0.1 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 0.1 mol/LKc = 0.08
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.1 mol/L solution of lactic acid (Ka = 0.00014).
c = 0.1 mol/LKa = 0.00014
pH =
12.pH & buffersHard
A 1.00 L buffer contains 0.25 mol hypochlorous acid and 0.15 mol of its conjugate base (Ka = 3 × 10⁻⁸). 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(CH4) = 24/16.04 = 1.496 mol
n(O2) = 20/32 = 0.625 mol
Limiting reactant
1.496/1 = 1.496, 0.625/2 = 0.3125 → O2
Mass of product
n(H2O) = 0.3125·2 = 0.625 mol
m = 0.625·18.02 = 11.26 g
Answer
m = 11.26 g
2.Stoichiometry
Answer: Y = 77.99 %
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(CaCO3) = 10/100.09 = 0.09991 mol
m = 0.09991·1/1·56.08 = 5.603 g
Percent yield
Y = 4.37/5.603·100 = 77.99%
Answer
Y = 77.99%
3.Molarity & dilution
Answer: Vwater = 100 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 = 0.4·500/0.5 = 400 mL
Water to add
Vwater = 500 − 400 = 100 mL
Answer
Vwater = 100 mL
4.Molarity & dilution
Answer: c = 1.514 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 = 0.2·300 + 2.5·400 = 1060 mmol
Concentration of the mixture
c = 1060/700 = 1.514 mol/L
Answer
c = 1.514 mol/L
5.Ideal gas law
Answer: P = 1103 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 210/44.01 = 4.772 mol
T = 5 + 273.15 = 278.1 K
Pressure
P = nRT/V = 4.772·8.314·278.1/0.01 = 1103 kPa
Answer
P = 1103 kPa
6.Ideal gas law
Answer: V2 = 1.411 L
Combined gas law: for a fixed amount of ideal gas, P1V1/T1 = P2V2/T2.
Absolute temperatures
T1 = 303.1 K, T2 = 513.1 K
Combined gas law
V2 = V1·(P1/P2)·(T2/T1) = 2·(250/600)·(513.1/303.1) = 1.411 L
Answer
V2 = 1.411 L
7.Calorimetry & enthalpy
Answer: Tf = 26.07 °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·(160 − Tf) = 300·4.18·(Tf − 24)
Solve for the final temperature
Tf = (19.4·160 + 1254·24)/(19.4 + 1254) = 26.07 °C
Answer
Tf = 26.07 °C
8.Calorimetry & enthalpy
Answer: q = 388.8 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
64 = (2x)²/(0.1 − x)²
Take the square root
8 = 2x/(0.1 − x) → x = 0.1·8/(2 + 8) = 0.08 mol/L
Equilibrium concentration
[H₂] = 0.1 − x = 0.02 mol/L
Answer
[H2] = 0.02 mol/L
10.Chemical equilibrium
Answer: [Cl2] = 0.05798 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/(0.1 − x) → x² + 0.08·x − 0.008 = 0
Solve the quadratic
x = (−0.08 + √(0.08² + 4·0.008))/2 = 0.05798 mol/L
Equilibrium concentration
[Cl₂] = x = 0.05798 mol/L
Answer
[Cl2] = 0.05798 mol/L
11.pH & buffers
Answer: pH = 2.435
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/(0.1 − x) → x = 0.003672 mol/L
pH
pH = −log(0.003672) = 2.435
Answer
pH = 2.435
12.pH & buffers
Answer: pH = 7.392
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⁻) = 0.15 + 0.02 = 0.17 mol
Henderson–Hasselbalch
pH = 7.523 + log(0.17/0.23) = 7.392
Answer
pH = 7.392
13.Reaction kinetics
Answer: t = 15.35 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)/0.3 = 15.35 min
Answer
t = 15.35 min
14.Reaction kinetics
Answer: T2 = 39.09 °C
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).