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Apply Amount of Chemical Change - IB Chemistry

  1. Consider two pathways to produce chloroethane, C₂H₅Cl. Pathway 1: C₂H₄ + HCl → C₂H₅Cl. Pathway 2: C₂H₆ + Cl₂ → C₂H₅Cl + HCl. Which statement correctly compares the atom economy of these two pathways?
    Answer: Pathway 1 has 100% atom economy because there is only one product. Atom economy is calculated as (Molar mass of desired product / Total molar mass of all reactants) × 100%. For Pathway 1, the only product is the desired product, C₂H₅Cl. Therefore, the mass of reactants equals the mass of the desired product, and the atom economy is 100%. For Pathway 2, HCl is a waste product, so the atom economy will be less than 100%. The physical state of a substance and the saturation of reactants do not directly determine atom economy. Percentage yield is an experimental measure and cannot be deduced from the balanced equation.
  2. 10.0 g of calcium nitrate, Ca(NO₃)₂, is heated strongly according to the equation: 2Ca(NO₃)₂(s) → 2CaO(s) + 4NO₂(g) + O₂(g). What is the total volume of gas produced, measured at STP? (Aᵣ: N=14.01, O=16.00, Ca=40.08. Molar volume of a gas at STP = 22.7 dm³ mol⁻¹)
    Answer: 3.46 dm³ First, calculate the molar mass of Ca(NO₃)₂: 40.08 + 2(14.01 + 3×16.00) = 164.10 g mol⁻¹. Then, find the moles of Ca(NO₃)₂: n = 10.0 g / 164.10 g mol⁻¹ = 0.06094 mol. The stoichiometry shows that for every 2 moles of Ca(NO₃)₂, a total of 4 + 1 = 5 moles of gas are produced. So, the mole ratio of reactant to total gas is 2:5. Moles of gas = 0.06094 mol Ca(NO₃)₂ × (5 mol gas / 2 mol Ca(NO₃)₂) = 0.1524 mol gas. Finally, calculate the volume at STP: Volume = 0.1524 mol × 22.7 dm³ mol⁻¹ = 3.46 dm³.
  3. In two separate experiments, a student combusts 1 mole of methane (CH₄) and 1 mole of methanol (CH₃OH). Which statement correctly compares the amount of oxygen required?
    Answer: Methanol requires less oxygen because it already contains an oxygen atom. Write balanced equations for complete combustion. Methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). 1 mole of CH₄ requires 2 moles of O₂. Methanol: 2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l). This means 1 mole of CH₃OH requires 1.5 moles of O₂. Therefore, methanol requires less oxygen. The reason is that the carbon in methanol is already partially oxidized, as indicated by the presence of the oxygen atom in the molecule, so less external oxygen is needed to complete the oxidation to CO₂.
  4. 25.0 cm³ of a sulfuric acid solution was diluted to 250.0 cm³. A 20.0 cm³ aliquot of this diluted solution required 18.5 cm³ of 0.100 mol dm⁻³ sodium hydroxide for complete neutralization. What was the concentration of the original, undiluted sulfuric acid solution?
    Answer: 0.463 mol dm⁻³ First, analyze the titration. The reaction is H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Moles of NaOH = C × V = 0.100 mol dm⁻³ × 0.0185 dm³ = 0.00185 mol. From the 1:2 mole ratio, moles of H₂SO₄ in the 20.0 cm³ aliquot = 0.00185 mol / 2 = 0.000925 mol. Now, find the concentration of the diluted acid. C = n / V = 0.000925 mol / 0.0200 dm³ = 0.04625 mol dm⁻³. This is the concentration of the diluted solution. The original solution was 10 times more concentrated (250.0 cm³ / 25.0 cm³ = 10). Therefore, the original concentration = 0.04625 mol dm⁻³ × 10 = 0.4625 mol dm⁻³. This is approximately 0.463 mol dm⁻³.
  5. A student adds 2.70 g of aluminium powder to 200 cm³ of 0.500 mol dm⁻³ copper(II) sulfate solution. The reaction is: 2Al(s) + 3CuSO₄(aq) → Al₂(SO₄)₃(aq) + 3Cu(s). What is the theoretical yield of copper in grams? (Aᵣ: Al=26.98, Cu=63.55)
    Answer: 6.36 g First, find the initial moles of each reactant. Moles Al = 2.70 g / 26.98 g mol⁻¹ = 0.100 mol. Moles CuSO₄ = concentration × volume = 0.500 mol dm⁻³ × 0.200 dm³ = 0.100 mol. Next, determine the limiting reactant using the mole ratio (2 Al : 3 CuSO₄). To react with 0.100 mol Al, we would need (3/2) × 0.100 = 0.150 mol CuSO₄. We only have 0.100 mol CuSO₄, so CuSO₄ is the limiting reactant. The amount of Cu produced is determined by the amount of CuSO₄. The ratio of CuSO₄ to Cu is 3:3 or 1:1. Therefore, 0.100 mol of Cu will be produced. Mass of Cu = moles × molar mass = 0.100 mol × 63.55 g mol⁻¹ = 6.36 g (rounded to 3 s.f.).
  6. Which statement best distinguishes between percentage yield and atom economy?
    Answer: Percentage yield compares the actual mass of product to the theoretical maximum, while atom economy measures the efficiency of reactant atom conversion into desired products. Percentage yield is a measure of experimental efficiency, calculated as (actual yield / theoretical yield) × 100%. It is affected by factors like incomplete reactions and loss of product during isolation. Atom economy is a theoretical measure of how efficiently atoms from the reactants are incorporated into the desired product, calculated from the balanced equation. It is independent of experimental conditions. Therefore, A is the correct distinction. B reverses the definitions. C is incorrect; a reaction can have a high yield but a low atom economy if it produces significant by-products. D is incorrect; atom economy is calculated from the equation, while percentage yield requires experimental data.
  7. A student performs a titration by adding 0.200 mol dm⁻³ HCl(aq) to 25.0 cm³ of 0.150 mol dm⁻³ Na₂CO₃(aq). The reaction is 2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂. What volume of HCl is required to reach the equivalence point?
    Answer: 37.5 cm³ First, calculate the moles of sodium carbonate in the flask. n(Na₂CO₃) = C × V = 0.150 mol dm⁻³ × 0.0250 dm³ = 0.00375 mol. According to the balanced equation, the mole ratio of HCl to Na₂CO₃ is 2:1. Therefore, the moles of HCl required is 2 × n(Na₂CO₃) = 2 × 0.00375 mol = 0.00750 mol. Finally, calculate the volume of HCl solution needed. V = n / C = 0.00750 mol / 0.200 mol dm⁻³ = 0.0375 dm³. To convert to cm³, multiply by 1000: 0.0375 dm³ × 1000 cm³/dm³ = 37.5 cm³.
  8. A buffer solution contains 0.25 M0.25 \text{ M} CH3COOH\text{CH}_3\text{COOH} and 0.15 M0.15 \text{ M} CH3COONa\text{CH}_3\text{COONa}. When 10.0 mL10.0 \text{ mL} of 0.20 M0.20 \text{ M} HCl\text{HCl} is added to 100.0 mL100.0 \text{ mL} of this buffer, what is the change in the amount (in moles) of the weak acid component?
    Answer: Increases by 0.0020 mol0.0020 \text{ mol} due to neutralization of conjugate base Initial moles of CH₃COOH = 0.100 L × 0.25 M = 0.025 mol. Moles HCl added = 0.010 L × 0.20 M = 0.0020 mol. The HCl reacts with acetate ion: CH₃COO⁻ + HCl → CH₃COOH + Cl⁻. This converts 0.0020 mol of acetate to 0.0020 mol of acetic acid, so acetic acid increases by 0.0020 mol. Choice B has wrong direction. Choice C uses wrong calculation. Choice D ignores the chemical change.
  9. For the reaction 2A + 5B → 3C, equal masses of reactants A and B are mixed. If the molar mass of A is twice the molar mass of B (Mₐ = 2Mₑ), which statement is correct?
    Answer: B is the limiting reactant because more moles of it are required by the stoichiometry. Let the mass of each reactant be 'm'. Moles of A = m / Mₐ. Moles of B = m / Mₑ. Since Mₐ = 2Mₑ, we can write Moles of A = m / (2Mₑ). Now let's find the ratio of moles available: (Moles of B) / (Moles of A) = (m / Mₑ) / (m / 2Mₑ) = 2. The stoichiometric ratio required for complete reaction is (Moles of B) / (Moles of A) = 5 / 2 = 2.5. Since the available ratio (2) is less than the required ratio (2.5), there is not enough B to react with all of A. Therefore, B is the limiting reactant.
  10. In a reaction vessel, 2.40 mol2.40 \text{ mol} of A\text{A} and 1.80 mol1.80 \text{ mol} of B\text{B} react according to 3A+2B→2C+D3\text{A} + 2\text{B} \rightarrow 2\text{C} + \text{D}. After the reaction reaches completion, 0.30 mol0.30 \text{ mol} of A\text{A} remains unreacted. What mass of product C\text{C} is formed if its molar mass is 64.0 g/mol64.0 \text{ g/mol}?
    Answer: 89.6 g89.6 \text{ g} based on the limiting reagent determination and stoichiometry If 0.30 mol A remains, then 2.40 - 0.30 = 2.10 mol A reacted. From stoichiometry: 2.10 mol A × (2 mol C/3 mol A) = 1.40 mol C formed. Mass = 1.40 mol × 64.0 g/mol = 89.6 g. Choice B incorrectly uses 1:1 stoichiometry (2.10 × 64.0/1.4). Choice C assumes B is limiting without checking (1.80 × 64.0). Choice D uses theoretical maximum without considering what actually reacted.
  11. 10 dm³ of nitrogen gas (N₂) and 25 dm³ of hydrogen gas (H₂) are mixed and reacted to form ammonia (NH₃) at a constant temperature and pressure: N₂(g) + 3H₂(g) → 2NH₃(g). What is the total volume of the gas mixture after the reaction is complete?
    Answer: 18.3 dm³ First, identify the limiting reactant. The stoichiometric ratio is 1 volume N₂ to 3 volumes H₂. To react with 10 dm³ of N₂, we would need 30 dm³ of H₂. Since we only have 25 dm³ of H₂, hydrogen is the limiting reactant. Now, calculate the volumes of gases reacting and produced based on the limiting reactant. Volume of H₂ reacted = 25 dm³. Volume of N₂ reacted = 25 dm³ / 3 = 8.33 dm³. Volume of NH₃ produced = 25 dm³ × (2/3) = 16.67 dm³. Finally, calculate the composition of the final mixture. Volume of unreacted N₂ = 10 dm³ - 8.33 dm³ = 1.67 dm³. Volume of unreacted H₂ = 0 dm³. Volume of NH₃ = 16.67 dm³. Total final volume = 1.67 dm³ + 16.67 dm³ = 18.34 dm³, which is approximately 18.3 dm³.
  12. In an electrolytic cell, a current of 2.50 A2.50 \text{ A} is passed through molten MgCl2\text{MgCl}_2 for 45.0 minutes45.0 \text{ minutes}. During this process, chlorine gas is evolved at the anode. What volume of Cl2\text{Cl}_2 gas is produced at STP?
    Answer: 0.785 L0.785 \text{ L} based on Faraday's laws and gas stoichiometry Charge = 2.50 A × 45.0 min × 60 s/min = 6750 C. Moles of electrons = 6750 C ÷ 96485 C/mol = 0.0700 mol e⁻. At anode: 2Cl⁻ → Cl₂ + 2e⁻, so moles Cl₂ = 0.0700 ÷ 2 = 0.0350 mol. Volume at STP = 0.0350 mol × 22.4 L/mol = 0.784 L ≈ 0.785 L. Choice B doesn't account for 2e⁻ per Cl₂. Choice C uses wrong conversion. Choice D multiplies by wrong factor.
  13. A 10.0 dm³ mixture of methane (CH₄) and ethane (C₂H₆) was completely combusted in excess oxygen, producing 16.0 dm³ of carbon dioxide. All gas volumes were measured at the same temperature and pressure. What was the volume of ethane in the original mixture?
    Answer: 6.0 dm³ Let V(CH₄) = x and V(C₂H₆) = y. So, x + y = 10.0. Write the combustion equations: CH₄ + 2O₂ → CO₂ + 2H₂O and 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O. By Avogadro's law, volume ratios equal mole ratios. From CH₄, volume of CO₂ produced is x. From C₂H₆, volume of CO₂ produced is 2y. Total volume of CO₂ is x + 2y = 16.0. Now we have a system of two linear equations: (1) x + y = 10.0 and (2) x + 2y = 16.0. Subtracting (1) from (2) gives y = 6.0 dm³. Therefore, the volume of ethane was 6.0 dm³.
  14. What mass of aluminium is produced when 1.0 tonne of aluminium oxide, Al₂O₃, is electrolyzed with a percentage yield of 90%? (1 tonne = 10⁶ g) (Aᵣ: O=16.00, Al=26.98)
    Answer: 0.48 tonnes The decomposition is 2Al₂O₃ → 4Al + 3O₂. M(Al₂O₃) = 2(26.98) + 3(16.00) = 101.96 g mol⁻¹. Mass of Al₂O₃ = 1.0 × 10⁶ g. Moles Al₂O₃ = 1.0 × 10⁶ g / 101.96 g mol⁻¹ = 9807.8 mol. From stoichiometry, moles of Al = 2 × moles of Al₂O₃ = 2 × 9807.8 = 19615.5 mol. Theoretical mass of Al = 19615.5 mol × 26.98 g mol⁻¹ = 529226 g = 0.529 tonnes. This is the theoretical yield. The actual yield is 90% of this value. Actual mass = 0.90 × 0.529 tonnes = 0.476 tonnes, which is approximately 0.48 tonnes.
  15. What volume of 0.500 mol dm⁻³ potassium hydroxide solution is required to completely neutralize 25.0 cm³ of 0.250 mol dm⁻³ phosphoric(V) acid, H₃PO₄? The reaction is: 3KOH + H₃PO₄ → K₃PO₄ + 3H₂O.
    Answer: 37.5 cm³ First, calculate the moles of phosphoric acid. n(H₃PO₄) = C × V = 0.250 mol dm⁻³ × 0.0250 dm³ = 0.00625 mol. From the balanced equation, the mole ratio of KOH to H₃PO₄ is 3:1. Therefore, the moles of KOH required is 3 × n(H₃PO₄) = 3 × 0.00625 mol = 0.01875 mol. Now, calculate the volume of KOH solution needed. V = n / C = 0.01875 mol / 0.500 mol dm⁻³ = 0.0375 dm³. Convert this volume to cm³: 0.0375 dm³ × 1000 cm³/dm³ = 37.5 cm³.
  16. A student mixes 50.0 mL50.0 \text{ mL} of 0.120 M0.120 \text{ M} AgNO3\text{AgNO}_3 with 30.0 mL30.0 \text{ mL} of 0.200 M0.200 \text{ M} NaCl\text{NaCl}. After the precipitation reaction is complete, the mixture is filtered. What is the concentration of Ag+\text{Ag}^+ ions remaining in the filtrate?
    Answer: 0.0000 M0.0000 \text{ M} because chloride is present in excess for complete precipitation Moles AgNO₃ = 0.0500 L × 0.120 M = 0.00600 mol Ag⁺. Moles NaCl = 0.0300 L × 0.200 M = 0.00600 mol Cl⁻. The reaction Ag⁺ + Cl⁻ → AgCl(s) has 1:1 stoichiometry, so both are completely consumed with no excess. [Ag⁺] = 0.0000 M. Choice B incorrectly assumes Ag⁺ excess. Choice C uses wrong calculation. Choice D ignores the precipitation completely.
  17. A 10.2 g sample of a metal oxide with the formula M₂O₃ was completely reduced, yielding 5.4 g of the metal M. What is the identity of metal M? (Aᵣ: O=16.00)
    Answer: Aluminium (Al) First, determine the mass of oxygen in the sample by subtraction: Mass of O = Mass of M₂O₃ - Mass of M = 10.2 g - 5.4 g = 4.8 g. Next, calculate the moles of oxygen atoms: n(O) = 4.8 g / 16.00 g mol⁻¹ = 0.30 mol. The formula M₂O₃ gives a mole ratio of M:O as 2:3. Use this ratio to find the moles of the metal M: n(M) = n(O) × (2/3) = 0.30 mol × (2/3) = 0.20 mol. Finally, calculate the molar mass of M: Molar Mass = mass / moles = 5.4 g / 0.20 mol = 27 g mol⁻¹. This molar mass corresponds to Aluminium (Al).
  18. When 1.50 g of zinc metal was added to excess hydrochloric acid, 496 cm³ of hydrogen gas was collected at STP. What is the percentage yield of this reaction? (Aᵣ: Zn=65.38. Molar volume of a gas at STP = 22.7 dm³ mol⁻¹)
    Answer: 95.2% The reaction is Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g). First, calculate the theoretical yield of H₂. Moles of Zn = 1.50 g / 65.38 g mol⁻¹ = 0.02294 mol. From the 1:1 stoichiometry, the theoretical moles of H₂ is also 0.02294 mol. Theoretical volume of H₂ = 0.02294 mol × 22.7 dm³ mol⁻¹ = 0.5208 dm³ = 520.8 cm³. The actual volume collected was 496 cm³. Percentage yield = (actual yield / theoretical yield) × 100% = (496 cm³ / 520.8 cm³) × 100% = 95.2%.
  19. A 4.97 g sample of hydrated copper(II) sulfate, CuSO₄·xH₂O, was heated to remove all water of crystallization. The mass of the anhydrous copper(II) sulfate, CuSO₄, was 3.19 g. What is the value of x? (Aᵣ: H=1.01, O=16.00, S=32.07, Cu=63.55)
    Answer: 5 First, find the mass of water lost: Mass H₂O = Mass of hydrate - Mass of anhydrous salt = 4.97 g - 3.19 g = 1.78 g. Next, convert the masses of the anhydrous salt and water to moles. M(CuSO₄) = 63.55 + 32.07 + 4(16.00) = 159.62 g mol⁻¹. Moles CuSO₄ = 3.19 g / 159.62 g mol⁻¹ = 0.0200 mol. M(H₂O) = 2(1.01) + 16.00 = 18.02 g mol⁻¹. Moles H₂O = 1.78 g / 18.02 g mol⁻¹ = 0.0988 mol. Finally, find the simplest whole number ratio of moles of H₂O to moles of CuSO₄. Ratio x = Moles H₂O / Moles CuSO₄ = 0.0988 mol / 0.0200 mol ≈ 4.94 ≈ 5. The value of x is 5.

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