30 g of H2 and 48 g of O2 react according to 2H2 + O2 → 2H2O. What mass m of H2O can form?
M(H2) = 2.016M(O2) = 32M(H2O) = 18.02 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 9.53 g of CaO is collected. Find the percent yield Y.
M(CaCO3) = 100.09 g/molM(CaO) = 56.08 g/molactual: 9.53 g
Y = %
3.Molarity & dilutionMedium
You need 500 mL of 0.25 mol/L solution and have a 4 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 4 mol/Lc2 = 0.25 mol/LV2 = 500 mL
V1 = mL
4.Molarity & dilutionHard
300 mL of a 1 mol/L solution is mixed with 100 mL of a 0.75 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
c1 = 1 mol/LV1 = 300 mLc2 = 0.75 mol/LV2 = 100 mL
c = mol/L
5.Ideal gas lawMedium
A tank of 20 L contains 380 g of CO₂ at 80 °C. Treat it as an ideal gas and find the pressure P.
A 200 g block of aluminium (c = 0.897 J/(g·°C)) at 180 °C is dropped into 300 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.
0.25 mol/L of H₂ and 0.25 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 25. Find the equilibrium concentration [H2].
[H₂]₀ = [I₂]₀ = 0.25 mol/LKc = 25
[H2] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.5 at the reaction temperature. A container starts with 0.8 mol/L of PCl₅ only. Find the equilibrium concentration [PCl5].
[PCl₅]₀ = 0.8 mol/LKc = 0.5
[PCl5] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.1 mol/L solution of hydrofluoric acid (Ka = 0.00068).
c = 0.1 mol/LKa = 0.00068
pH =
12.pH & buffersHard
A buffer contains 0.4 mol/L formic acid and 0.1 mol/L of its sodium salt. Ka = 0.00018. Find the pH pH of the buffer.
[HA] = 0.4[A⁻] = 0.1Ka = 0.00018
pH =
13.Reaction kineticsMedium
A reaction with rate law −rA = k·CA² has k = 1 L/(mol·min) and starts at 0.5 mol/L. Find the concentration CA after 20 min.
k = 1 L/(mol·min)CA0 = 0.5 mol/Lt = 20 min
CA = mol/L
14.Reaction kineticsHard
A first-order reaction has a rate constant of 0.005 s⁻¹ at 30 °C and an activation energy of 40 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(H2) = 30/2.016 = 14.88 mol
n(O2) = 48/32 = 1.5 mol
Limiting reactant
14.88/2 = 7.44, 1.5/1 = 1.5 → O2
Mass of product
n(H2O) = 1.5·2 = 3 mol
m = 3·18.02 = 54.06 g
Answer
m = 54.06 g
2.Stoichiometry
Answer: Y = 85.04 %
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 = 9.53/11.21·100 = 85.04%
Answer
Y = 85.04%
3.Molarity & dilution
Answer: V1 = 31.25 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 = 0.25·500/4 = 31.25 mL
Answer
V1 = 31.25 mL
4.Molarity & dilution
Answer: c = 0.9375 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 = 1·300 + 0.75·100 = 375 mmol
Concentration of the mixture
c = 375/400 = 0.9375 mol/L
Answer
c = 0.9375 mol/L
5.Ideal gas law
Answer: P = 1268 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 380/44.01 = 8.634 mol
T = 80 + 273.15 = 353.1 K
Pressure
P = nRT/V = 8.634·8.314·353.1/0.02 = 1268 kPa
Answer
P = 1268 kPa
6.Ideal gas law
Answer: V2 = 3.385 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·(350/350)·(513.1/303.1) = 3.385 L
Answer
V2 = 3.385 L
7.Calorimetry & enthalpy
Answer: Tf = 41.77 °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
200·0.897·(180 − Tf) = 300·4.18·(Tf − 22)
Solve for the final temperature
Tf = (179.4·180 + 1254·22)/(179.4 + 1254) = 41.77 °C
Answer
Tf = 41.77 °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.
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
0.5 = x²/(0.8 − x) → x² + 0.5·x − 0.4 = 0
Solve the quadratic
x = (−0.5 + √(0.5² + 4·0.4))/2 = 0.4301 mol/L
Equilibrium concentration
[PCl₅] = 0.8 − 0.4301 = 0.3699 mol/L
Answer
[PCl5] = 0.3699 mol/L
11.pH & buffers
Answer: pH = 2.102
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 = 0.007913 mol/L
pH
pH = −log(0.007913) = 2.102
Answer
pH = 2.102
12.pH & buffers
Answer: pH = 3.143
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 = 3.745
Henderson–Hasselbalch
pH = 3.745 + log(0.1/0.4) = 3.143
Answer
pH = 3.143
13.Reaction kinetics
Answer: CA = 0.04545 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 = 2 + 1·20 = 22 L/mol
CA = 1/22 = 0.04545 mol/L
Answer
CA = 0.04545 mol/L
14.Reaction kinetics
Answer: A = 39036 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).