These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: H 1.008, N 14.01, O 16.00, Mg 24.31, Al 26.98, P 30.97, Cl 35.45, Ca 40.08, Cu 63.55 g mol⁻¹; R = 0.08206 L atm mol⁻¹ K⁻¹; T(K) = T(°C) + 273.15. Assume every reaction goes to completion unless the question says otherwise.
Question 1 (multiple choice · foundation)
Aluminium powder burns in oxygen: 4Al(s) + 3O₂(g) → 2Al₂O₃(s). How many moles of O₂ react with 0.60 mol of Al?
- (A) 0.30 mol
- (B) 0.45 mol
- (C) 0.60 mol
- (D) 0.80 mol
Answer and explanation
Answer: (B). n(O₂) = 0.60 mol Al × (3 mol O₂ / 4 mol Al) = 0.45 mol.
- (A) uses the ratio of Al to Al₂O₃ (2/4) instead of O₂ to Al.
- (C) assumes a 1:1 ratio and ignores the coefficients.
- (D) turns the ratio upside down (4/3). The units would not cancel: you would be left with mol² of Al on top.
Question 2 (multiple choice · core)
A student mixes 14.0 g of N₂ with 4.00 g of H₂, which react as N₂(g) + 3H₂(g) → 2NH₃(g). Which statement is correct?
- (A) H₂ is limiting, and the maximum mass of NH₃ is 22.5 g.
- (B) N₂ is limiting, and the maximum mass of NH₃ is 17.0 g.
- (C) N₂ is limiting, and the maximum mass of NH₃ is 8.51 g.
- (D) H₂ is limiting, and the maximum mass of NH₃ is 33.8 g.
Answer and explanation
Answer: (B). n(N₂) = 14.0 ÷ 28.02 = 0.4996 mol; n(H₂) = 4.00 ÷ 2.016 = 1.984 mol. Divide by coefficients: N₂ 0.4996 ÷ 1 = 0.4996; H₂ 1.984 ÷ 3 = 0.661. N₂ gives the smaller value, so N₂ is limiting. n(NH₃) = 0.4996 × 2 = 0.9993 mol, and M(NH₃) = 17.03 g mol⁻¹, so the mass is 17.0 g.
- (A) picks H₂ because its mass is smaller. That compares grams, not moles. 22.5 g is the NH₃ that H₂ could make if N₂ never ran out.
- (C) has the right limiting reactant but forgets the 2 in front of NH₃.
- (D) picks H₂ and also uses a 1:1 ratio. 33.8 g of product from 18.0 g of reactants would break conservation of mass.
Question 3 (multiple choice · core)
A student adds 0.50 g of magnesium ribbon to 50.0 mL of 2.0 M HCl and collects the hydrogen: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). All of the magnesium dissolves. She repeats the experiment with the same mass of magnesium but 100.0 mL of the same acid, at the same temperature and pressure. How does the volume of hydrogen compare with the first experiment?
- (A) It is half as large.
- (B) It is the same.
- (C) It is twice as large.
- (D) It is four times as large.
Answer and explanation
Answer: (B). n(Mg) = 0.50 ÷ 24.31 = 0.0206 mol, which needs 0.0411 mol HCl. The first experiment already has 0.0500 L × 2.0 M = 0.100 mol HCl, so the acid is in excess and Mg is limiting. Doubling the acid only adds more excess. The hydrogen is still set by the magnesium.
- (A) has the relationship backwards; more acid cannot make less gas.
- (C) assumes the gas is proportional to the acid, which is only true when the acid is limiting.
- (D) squares the change for no chemical reason (perhaps from the coefficient 2).
Question 4 (calculation · core)
A technician removes calcium ions from 40.0 mL of 0.225 M CaCl₂ solution by adding 0.150 M sodium phosphate:
3CaCl₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6NaCl(aq)
(a) What volume of Na₃PO₄ solution is needed to react exactly with all the CaCl₂? (b) What mass of calcium phosphate forms?
Worked solution
- n(CaCl₂) = 0.225 mol L⁻¹ × 0.0400 L = 0.00900 mol.
- n(Na₃PO₄) = 0.00900 mol × (2 / 3) = 0.00600 mol.
- V = n / M = 0.00600 mol ÷ 0.150 mol L⁻¹ = 0.0400 L = 40.0 mL.
- n(Ca₃(PO₄)₂) = 0.00900 mol CaCl₂ × (1 / 3) = 0.00300 mol. M = 3(40.08) + 2(30.97) + 8(16.00) = 310.18 g mol⁻¹. Mass = 0.00300 × 310.18 = 0.931 g.
Suggested mark points (3): 1 for moles of CaCl₂ using 0.0400 L; 1 for the 2/3 ratio and 40.0 mL; 1 for 0.931 g with the 1/3 ratio and correct molar mass.
Common error: using a 1:1 ratio gives 60.0 mL, which would leave phosphate in excess.
Question 5 (calculation · core)
A 20.0 mL sample of 0.880 M hydrogen peroxide solution decomposes completely in the presence of a catalyst:
2H₂O₂(aq) → 2H₂O(l) + O₂(g)
What volume of oxygen gas forms at 22 °C and 0.975 atm?
Worked solution
- n(H₂O₂) = 0.880 mol L⁻¹ × 0.0200 L = 0.0176 mol.
- n(O₂) = 0.0176 mol × (1 / 2) = 0.00880 mol.
- T = 22 + 273.15 = 295.15 K.
- V = nRT / P = (0.00880)(0.08206)(295.15) ÷ 0.975 = 0.2186 L = 0.219 L (219 mL).
Suggested mark points (3): 1 for moles of H₂O₂ with volume in litres; 1 for the 1:2 ratio; 1 for the volume with T in kelvin and correct units.
Common error: using T = 22 instead of 295.15 gives 16.3 mL, far too small for this amount of gas.
Question 6 (constructed response · core)
A student places 3.00 g of aluminium foil in 50.0 mL of 1.50 M copper(II) sulfate solution. Copper metal forms:
2Al(s) + 3CuSO₄(aq) → Al₂(SO₄)₃(aq) + 3Cu(s)
(a) Identify the limiting reactant. Show your reasoning. (b) Calculate the theoretical yield of copper, in grams. (c) Calculate the mass of the excess reactant left when the reaction is complete. (d) The student collects 4.31 g of dry copper. Calculate the percent yield. (e) The student repeats the experiment with 6.00 g of aluminium and the same copper(II) sulfate solution. Predict the mass of copper that forms, and justify your prediction.
Worked solution and suggested Marlbridge rubric
(a) n(Al) = 3.00 ÷ 26.98 = 0.11119 mol. n(CuSO₄) = 1.50 × 0.0500 = 0.0750 mol. Divide by coefficients: Al 0.11119 ÷ 2 = 0.0556; CuSO₄ 0.0750 ÷ 3 = 0.0250. CuSO₄ is limiting. (Equivalently, 0.0750 mol CuSO₄ needs only 0.0500 mol Al, and 0.111 mol is present.)
(b) n(Cu) = n(CuSO₄) = 0.0750 mol. Mass = 0.0750 × 63.55 = 4.77 g.
(c) Al used = 0.0750 × (2 / 3) = 0.0500 mol. Al left = 0.11119 − 0.0500 = 0.06119 mol = 0.06119 × 26.98 = 1.65 g.
(d) Percent yield = 4.31 ÷ 4.7663 × 100 = 90.4%.
(e) Still 4.77 g. Aluminium was already in excess, so CuSO₄ is still limiting. Doubling the aluminium only increases the aluminium left over; it cannot make more copper than 0.0750 mol of Cu²⁺ provides.
| Point | What earns it |
|---|---|
| 1 | Moles of both reactants, with the volume in litres |
| 1 | Correct limiting reactant, justified by comparing moles with the coefficients (not masses) |
| 1 | Theoretical yield 4.77 g from the limiting reactant |
| 1 | Excess Al 1.65 g, using the 2:3 ratio |
| 1 | Percent yield 90.4% (accept a value carried forward from (b)) |
| 1 | Prediction of no change with the reason that Al is in excess and CuSO₄ still limits |
If (a) is wrong, award (b) to (d) for correct working carried forward. Taking Al as limiting gives 10.6 g of copper, more than twice the mass of copper in the solution; this should alert you to the error.
Question 7 (explanation · stretch)
A sealed container holds 6 molecules of NO and 4 molecules of O₂. They react completely according to 2NO(g) + O₂(g) → 2NO₂(g).
(a) Identify the limiting reactant. (b) Describe the contents of the container after the reaction, giving the number of each kind of molecule. (c) Show that the number of each kind of atom is the same before and after. (d) A second container starts with 6 NO and 6 O₂. A third starts with 8 NO and 4 O₂. Predict the number of NO₂ molecules formed in each, and explain the difference using the idea of a limiting reactant.
Model answer and suggested Marlbridge rubric
(a) 6 NO need 3 O₂, and 4 O₂ are present. NO is limiting.
(b) All 6 NO react with 3 O₂ to make 6 NO₂, leaving 1 O₂ unreacted. No NO remains.
(c) Before: N = 6; O = 6 + (4 × 2) = 14. After: N = 6 (in NO₂); O = (6 × 2) + (1 × 2) = 14. Atoms are conserved.
(d) Container 2: still 6 NO₂ (3 O₂ left over). NO was already limiting, so extra O₂ is only extra excess. Container 3: 8 NO₂ with no O₂ left. Adding NO increases the limiting reactant, so more product forms; 8 NO need exactly 4 O₂, so now both reactants are used up.
| Point | What earns it |
|---|---|
| 1 | NO identified as limiting, with the 2:1 reasoning |
| 1 | 6 NO₂ and 1 O₂ after the reaction |
| 1 | Correct atom count showing conservation of N and O |
| 1 | Both predictions (6 and 8 NO₂) with the explanation that only extra limiting reactant increases product |
How did you do?
- Q1 or Q4 wrong: re-read “Why a balanced equation lets you predict amounts” in the study guide and check you put the wanted substance on top of the ratio.
- Q2 or Q6 wrong: practise the three-step limiting-reactant method in Worked example 2.
- Q5 wrong: check your unit conversions (mL to L, °C to K) against Figure 1.
- Q3 or Q7 incomplete: use Figure 2 to see why extra excess reactant changes nothing.
Then tick off the topic checklist.