This checkpoint covers Lessons 11–16: stoichiometry and mole ratios, mass–mass calculations, limiting reagents, percentage yield and purity, gas stoichiometry and the gas laws, and solution stoichiometry. 20 multiple choice + 5 short answer questions.
Lesson Summaries, Quick Review
A balanced equation's coefficients give the mole ratio in which substances react and form. To find the moles of a target substance: write the balanced equation, find moles of the known substance, then multiply by the mole ratio (target ÷ known).
Four steps: (1) balance the equation, (2) convert the known mass to moles (n = m ÷ MM), (3) apply the mole ratio to get moles of target, (4) convert moles of target to mass (m = n × MM). The mole ratio always comes from the coefficients, not the masses.
The limiting reagent is fully consumed and determines the maximum product. Identify it by dividing each reactant's moles by its coefficient; the smallest value is limiting. Always base the product calculation on the limiting reagent; the other reactant is in excess.
% yield = (actual ÷ theoretical) × 100; the true yield is ≤ 100%, but a measured value can exceed 100% if the product is wet or impure (an error signal). % purity = (mass of pure substance ÷ mass of sample) × 100. Apply purity before stoichiometry; apply yield at the end.
Convert gas volumes to moles with n = V ÷ Vm (STP 22.71, SATP 24.8 L mol⁻¹). For changing conditions use the gas laws (T in kelvin): Boyle P₁V₁ = P₂V₂; Charles V₁/T₁ = V₂/T₂; combined P₁V₁/T₁ = P₂V₂/T₂; ideal gas law PV = nRT (R = 8.314). Solution stoichiometry (L16) uses n = cV in the same way.
Updates as you answer. SA marks are self-assessed from model answers.
1. In N₂ + 3H₂ → 2NH₃, how many moles of H₂ react completely with 2.0 mol of N₂? L11
2. For 2Al + 3Cl₂ → 2AlCl₃, what is the mole ratio of Al to AlCl₃? L11
3. For C₃H₈ + 5O₂ → 3CO₂ + 4H₂O, how many moles of CO₂ form from 0.40 mol of C₃H₈? L11
4. Why are the coefficients of a balanced equation called the mole ratio? L11
5. When given the mass of a reactant, what is the correct first step in a mass–mass problem? L12
6. For 2H₂ + O₂ → 2H₂O, what mass of water forms from 4.00 g of H₂? (H₂ = 2.016, H₂O = 18.0) L12
7. For CaCO₃ → CaO + CO₂, what mass of CaO forms from 50.0 g of CaCO₃? (CaCO₃ = 100.09, CaO = 56.08) L12
8. In a mass–mass calculation, the mole ratio is taken from: L12
9. In Fe + S → FeS, 0.20 mol of Fe is mixed with 0.25 mol of S. Which is the limiting reagent? L13
10. In 2Na + Cl₂ → 2NaCl, 4.0 mol of Na is mixed with 1.5 mol of Cl₂. Which is limiting? L13
11. How is the limiting reagent identified when the moles of each reactant are known? L13
12. When a reaction goes to completion, the limiting reagent is: L13
13. A reaction has a theoretical yield of 25.0 g and an actual yield of 19.0 g. What is the percentage yield? L14
14. A student measures an apparent percentage yield above 100%. The most likely cause is: L14
15. A 5.00 g ore sample is 80.0% Fe₂O₃ by mass. What mass of pure Fe₂O₃ does it contain? L14
16. Percentage purity is calculated as: L14
17. What volume does 0.250 mol of CO₂ occupy at STP? (Vm = 22.71 L mol⁻¹) L15
18. A gas occupies 2.0 L at 100 kPa. At constant temperature, what volume does it occupy at 200 kPa? L15
19. In any gas-law calculation, temperature must be expressed in: L15
20. Using PV = nRT (R = 8.314), what is the pressure of 0.50 mol of gas at 300 K in a 5.0 L container? L15
Attempt all five questions before checking answers.
21. For 4NH₃ + 5O₂ → 4NO + 6H₂O, calculate the moles of NO formed when 2.0 mol of NH₃ reacts completely. 3 MARKS
22. Magnesium burns in oxygen: 2Mg + O₂ → 2MgO. Calculate the mass of MgO formed from 6.08 g of magnesium. (Mg = 24.31, MgO = 40.30) 4 MARKS
23. For N₂ + 3H₂ → 2NH₃, 28.0 g of N₂ reacts with 9.00 g of H₂. (a) Identify the limiting reagent (show the comparison). (b) Calculate the theoretical mass of NH₃. (N = 14.01, H = 1.008) 4 MARKS
24. A 10.0 g sample of impure calcium carbonate is fully decomposed and produces 3.74 g of CO₂. Calculate the percentage purity of the CaCO₃. (CaCO₃ = 100.09, CO₂ = 44.01) 3 MARKS
25. (a) A gas occupies 250 mL at 300 K and 100 kPa. Calculate its volume at 350 K and 100 kPa (constant pressure). (b) 25.0 mL of 0.100 mol L⁻¹ AgNO₃ reacts with excess NaCl: AgNO₃ + NaCl → AgCl↓ + NaNO₃. Calculate the mass of AgCl precipitate. (AgCl = 143.32) 4 MARKS
1. A Ratio N₂:H₂ = 1:3, so 2.0 mol N₂ needs 2.0 × 3 = 6.0 mol H₂.
2. A Coefficients Al : AlCl₃ = 2 : 2 = 1 : 1.
3. A Ratio C₃H₈:CO₂ = 1:3, so 0.40 × 3 = 1.2 mol CO₂.
4. B Coefficients give the ratio of moles in which substances react and form (from conservation of atoms).
5. B Always convert the known mass to moles first (n = m ÷ MM), then apply the mole ratio.
6. A n(H₂) = 4.00 ÷ 2.016 = 1.984 mol; ratio H₂:H₂O = 1:1; m(H₂O) = 1.984 × 18.0 = 35.7 g.
7. A n(CaCO₃) = 50.0 ÷ 100.09 = 0.4996 mol; n(CaO) = 0.4996; m = 0.4996 × 56.08 = 28.0 g.
8. C The mole ratio always comes from the coefficients of the balanced equation.
9. A 1:1 ratio; Fe (0.20 mol) < S (0.25 mol), so Fe is limiting.
10. A Na: 4.0 ÷ 2 = 2.0; Cl₂: 1.5 ÷ 1 = 1.5. Cl₂ has the smaller value, so Cl₂ is limiting.
11. B Divide each reactant's moles by its coefficient; the smallest quotient is the limiting reagent.
12. B The limiting reagent is completely consumed; the other reactant is left in excess.
13. A % yield = (19.0 ÷ 25.0) × 100 = 76.0%.
14. B The true yield cannot exceed 100%; an apparent value above 100% means the product was wet or impure (inflated mass).
15. A m(pure) = 5.00 × (80.0 ÷ 100) = 4.00 g.
16. B % purity = (mass of pure substance ÷ mass of sample) × 100.
17. A V = n × Vm = 0.250 × 22.71 = 5.68 L at STP.
18. A Boyle's law: P₁V₁ = P₂V₂ → V₂ = (100 × 2.0) ÷ 200 = 1.0 L.
19. B All gas-law temperatures must be in kelvin (T(K) = T(°C) + 273.15).
20. A P = nRT ÷ V = (0.50 × 8.314 × 300) ÷ 5.0 = 249 kPa.
Q21 (3 marks): Ratio NH₃:NO = 4:4 = 1:1 [1]; so n(NO) = n(NH₃) = 2.0 mol [2].
Q22 (4 marks): n(Mg) = 6.08 ÷ 24.31 = 0.2501 mol [1]; ratio Mg:MgO = 2:2 = 1:1, so n(MgO) = 0.2501 mol [1]; m = 0.2501 × 40.30 = 10.1 g [2].
Q23 (4 marks): (a) n(N₂) = 28.0 ÷ 28.02 = 0.999 mol (÷1 = 0.999); n(H₂) = 9.00 ÷ 2.016 = 4.46 mol (÷3 = 1.49). N₂ has the smaller value → N₂ is limiting [2]. (b) n(NH₃) = 2 × 0.999 = 1.999 mol; MM(NH₃) = 17.03; m = 1.999 × 17.03 = 34.0 g [2].
Q24 (3 marks): n(CO₂) = 3.74 ÷ 44.01 = 0.08498 mol [1]; n(CaCO₃) = 0.08498 mol (1:1); m(CaCO₃) = 0.08498 × 100.09 = 8.51 g [1]; % purity = (8.51 ÷ 10.0) × 100 = 85.1% [1].
Q25 (4 marks): (a) Charles' law: V₂ = V₁ × T₂ ÷ T₁ = 250 × 350 ÷ 300 = 292 mL [2]. (b) n(AgNO₃) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol; n(AgCl) = 2.50 × 10⁻³ mol (1:1); m = 2.50 × 10⁻³ × 143.32 = 0.358 g [2].
Tick when you have checked all answers and noted areas to review.