20 g of CH4 and 64 g of O2 react according to CH4 + 2O2 → CO2 + 2H2O. What mass m of CO2 can form?
M(CH4) = 16.04M(O2) = 32M(CO2) = 44.01 g/mol
m = g
2.StoichiometryHard
In the reaction CaCO3 → CaO + CO2, 24 g of CaCO3 reacts completely (the other reactants are in excess) and 8.23 g of CO2 is collected. Find the percent yield Y.
M(CaCO3) = 100.09 g/molM(CO2) = 44.01 g/molactual: 8.23 g
Y = %
3.Molarity & dilutionMedium
You need 100 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 = 100 mL
V1 = mL
4.Molarity & dilutionHard
100 mL of a 0.5 mol/L solution is mixed with 250 mL of a 0.6 mol/L solution of the same substance. Find the concentration c of the mixture (volumes are additive).
A 50 g block of aluminium (c = 0.897 J/(g·°C)) at 150 °C is dropped into 400 g of water (c = 4.18 J/(g·°C)) at 16 °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 C4H10 (M = 58.12 g/mol) burns completely: 2C4H10 + 13O₂ → 8CO₂ + 10H₂O(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 = 36. Find the equilibrium concentration [HI].
[H₂]₀ = [I₂]₀ = 1 mol/LKc = 36
[HI] = mol/L
10.Chemical equilibriumHard
PCl₅ decomposes: PCl₅ ⇌ PCl₃ + Cl₂, with Kc = 0.02 at the reaction temperature. A container starts with 0.2 mol/L of PCl₅ only. Find the equilibrium concentration [Cl2].
[PCl₅]₀ = 0.2 mol/LKc = 0.02
[Cl2] = mol/L
11.pH & buffersMedium
Find the pH pH of a 0.25 mol/L solution of formic acid (Ka = 0.00018).
c = 0.25 mol/LKa = 0.00018
pH =
12.pH & buffersHard
A 1.00 L buffer contains 0.3 mol hypochlorous acid and 0.1 mol of its conjugate base (Ka = 3 × 10⁻⁸). 0.025 mol of NaOH is added. Find the new pH pH.
A zero-order reaction has k = 0.025 mol/(L·min) and starts at 2 mol/L. How long does it take to reach 80% conversion? Find t.
k = 0.025 mol/(L·min)CA0 = 2 mol/LX = 80 %
t = min
14.Reaction kineticsHard
A first-order reaction has a rate constant of 0.1 s⁻¹ at 25 °C and an activation energy of 80 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(CH4) = 20/16.04 = 1.247 mol
n(O2) = 64/32 = 2 mol
Limiting reactant
1.247/1 = 1.247, 2/2 = 1 → O2
Mass of product
n(CO2) = 1·1 = 1 mol
m = 1·44.01 = 44.01 g
Answer
m = 44.01 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) = 24/100.09 = 0.2398 mol
m = 0.2398·1/1·44.01 = 10.55 g
Percent yield
Y = 8.23/10.55·100 = 77.99%
Answer
Y = 77.99%
3.Molarity & dilution
Answer: V1 = 12 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 = 0.3·100/2.5 = 12 mL
Answer
V1 = 12 mL
4.Molarity & dilution
Answer: c = 0.5714 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 = 0.5·100 + 0.6·250 = 200 mmol
Concentration of the mixture
c = 200/350 = 0.5714 mol/L
Answer
c = 0.5714 mol/L
5.Ideal gas law
Answer: V = 2420 L
Ideal gas law: PV = nRT, with n = m/M.
Moles
n = 330/4.003 = 82.44 mol
T = 80 + 273.15 = 353.1 K
Volume
V = nRT/P = 82.44·8.314·353.1/100 = 2420 L
Answer
V = 2420 L
6.Ideal gas law
Answer: ρ = 0.7123 kg/m³
Ideal gas density: ρ = PM/(RT).
Rearrange the gas law
PV = (m/M)RT → ρ = m/V = PM/(RT)
T = 200 + 273.15 = 473.1 K
Density
ρ = 700·4.003/(8.314·473.1) = 0.7123 kg/m³
Answer
ρ = 0.7123 kg/m³
7.Calorimetry & enthalpy
Answer: Tf = 19.5 °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
50·0.897·(150 − Tf) = 400·4.18·(Tf − 16)
Solve for the final temperature
Tf = (44.85·150 + 1672·16)/(44.85 + 1672) = 19.5 °C
Answer
Tf = 19.5 °C
8.Calorimetry & enthalpy
Answer: q = 1238 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
36 = (2x)²/(1 − x)²
Take the square root
6 = 2x/(1 − x) → x = 1·6/(2 + 6) = 0.75 mol/L
Equilibrium concentration
[HI] = 2x = 1.5 mol/L
Answer
[HI] = 1.5 mol/L
10.Chemical equilibrium
Answer: [Cl2] = 0.05403 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
0.02 = x²/(0.2 − x) → x² + 0.02·x − 0.004 = 0
Solve the quadratic
x = (−0.02 + √(0.02² + 4·0.004))/2 = 0.05403 mol/L
Equilibrium concentration
[Cl₂] = x = 0.05403 mol/L
Answer
[Cl2] = 0.05403 mol/L
11.pH & buffers
Answer: pH = 2.179
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/(0.25 − x) → x = 0.006619 mol/L
pH
pH = −log(0.006619) = 2.179
Answer
pH = 2.179
12.pH & buffers
Answer: pH = 7.18
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) = 0.3 − 0.025 = 0.275 mol
n(A⁻) = 0.1 + 0.025 = 0.125 mol
Henderson–Hasselbalch
pH = 7.523 + log(0.125/0.275) = 7.18
Answer
pH = 7.18
13.Reaction kinetics
Answer: t = 64 min
Integrated rate laws: zero order: C = C₀ − kt (straight line, t½ = C₀/(2k)); first order: ln(C₀/C) = kt (t½ = ln 2/k); second order: 1/C − 1/C₀ = kt (t½ = 1/(kC₀)). Conversion X means C = C₀(1 − X).
Amount converted
CA0·X = 2·0.8 = 1.6 mol/L
Time
t = CA0·X/k = 1.6/0.025 = 64 min
Answer
t = 64 min
14.Reaction kinetics
Answer: A = 1.038 × 10¹³ 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).