6 g of Na and 16 g of Cl2 react according to 2Na + Cl2 → 2NaCl. What mass m of NaCl can form?
M(Na) = 22.99M(Cl2) = 70.9M(NaCl) = 58.44 g/mol
m = g
2.StoichiometryHard
In the reaction 2Mg + O2 → 2MgO, 10 g of Mg reacts completely (the other reactants are in excess) and 14.59 g of MgO is collected. Find the percent yield Y.
M(Mg) = 24.31 g/molM(MgO) = 40.3 g/molactual: 14.59 g
Y = %
3.Molarity & dilutionMedium
You need 500 mL of 0.3 mol/L solution and have a 2.5 mol/L stock solution. What volume V1 of stock solution do you need?
c1 = 2.5 mol/Lc2 = 0.3 mol/LV2 = 500 mL
V1 = mL
4.Molarity & dilutionHard
150 mL of a 0.2 mol/L solution is mixed with 50 mL of a 0.1 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
c1 = 0.2 mol/LV1 = 150 mLc2 = 0.1 mol/LV2 = 50 mL
c = mol/L
5.Ideal gas lawMedium
A tank of 5 L contains 40 g of N₂ at 30 °C. Treat it as an ideal gas and find the pressure P.
A 75 g block of zinc (c = 0.388 J/(g·°C)) at 130 °C is dropped into 400 g of water (c = 4.18 J/(g·°C)) at 24 °C in an insulated cup. Find the final temperature Tf.
0.2 mol/L of H₂ and 0.2 mol/L of I₂ are put in a container and react: H₂ + I₂ ⇌ 2HI, with Kc = 50. Find the equilibrium concentration [HI].
[H₂]₀ = [I₂]₀ = 0.2 mol/LKc = 50
[HI] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.1 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 = 0.1
[PCl5] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.01 mol/L solution of formic acid (Ka = 0.00018).
c = 0.01 mol/LKa = 0.00018
pH =
12.pH & buffersHard
A 1.00 L buffer contains 0.1 mol formic acid and 0.1 mol of its conjugate base (Ka = 0.00018). 0.025 mol of NaOH is added. Find the new pH pH.
A reaction with rate law −rA = k·CA² has k = 0.8 L/(mol·min) and starts at 1.5 mol/L. Find the concentration CA after 15 min.
k = 0.8 L/(mol·min)CA0 = 1.5 mol/Lt = 15 min
CA = mol/L
14.Reaction kineticsHard
A first-order reaction has a rate constant of 0.001 s⁻¹ at 100 °C and an activation energy of 50 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(Na) = 6/22.99 = 0.261 mol
n(Cl2) = 16/70.9 = 0.2257 mol
Limiting reactant
0.261/2 = 0.1305, 0.2257/1 = 0.2257 → Na
Mass of product
n(NaCl) = 0.1305·2 = 0.261 mol
m = 0.261·58.44 = 15.25 g
Answer
m = 15.25 g
2.Stoichiometry
Answer: Y = 88.01 %
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(Mg) = 10/24.31 = 0.4114 mol
m = 0.4114·2/2·40.3 = 16.58 g
Percent yield
Y = 14.59/16.58·100 = 88.01%
Answer
Y = 88.01%
3.Molarity & dilution
Answer: V1 = 60 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 = 0.3·500/2.5 = 60 mL
Answer
V1 = 60 mL
4.Molarity & dilution
Answer: c = 0.175 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 = 0.2·150 + 0.1·50 = 35 mmol
Concentration of the mixture
c = 35/200 = 0.175 mol/L
Answer
c = 0.175 mol/L
5.Ideal gas law
Answer: P = 719.9 kPa
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = m/M = 40/28.01 = 1.428 mol
T = 30 + 273.15 = 303.1 K
Pressure
P = nRT/V = 1.428·8.314·303.1/0.005 = 719.9 kPa
Answer
P = 719.9 kPa
6.Ideal gas law
Answer: ρ = 1.49 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 50 + 273.15 = 323.1 K
Density
ρ = 1000·4.003/(8.314·323.1) = 1.49 kg/m³
Answer
ρ = 1.49 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 25.81 °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
75·0.388·(130 − Tf) = 400·4.18·(Tf − 24)
Solve for the final temperature
Tf = (29.1·130 + 1672·24)/(29.1 + 1672) = 25.81 °C
Answer
Tf = 25.81 °C
8.Calorimetry & enthalpy
Answer: ΔH° = −2220 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] = 0.3118 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.
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/(0.8 − x) → x² + 0.1·x − 0.08 = 0
Solve the quadratic
x = (−0.1 + √(0.1² + 4·0.08))/2 = 0.2372 mol/L
Equilibrium concentration
[PCl₅] = 0.8 − 0.2372 = 0.5628 mol/L
Answer
[PCl5] = 0.5628 mol/L
11.pH & buffers
Answer: pH = 2.901
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 = 0.001255 mol/L
pH
pH = −log(0.001255) = 2.901
Answer
pH = 2.901
12.pH & buffers
Answer: pH = 3.967
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) = 0.1 − 0.025 = 0.075 mol
n(A⁻) = 0.1 + 0.025 = 0.125 mol
Henderson–Hasselbalch
pH = 3.745 + log(0.125/0.075) = 3.967
Answer
pH = 3.967
13.Reaction kinetics
Answer: CA = 0.07895 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 = 0.6667 + 0.8·15 = 12.67 L/mol
CA = 1/12.67 = 0.07895 mol/L
Answer
CA = 0.07895 mol/L
14.Reaction kinetics
Answer: A = 9986 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).