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A Level Chemistry H3 Stoichiometry Moles Quiz

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Questions

A-Level Chemistry H3 Quiz - Stoichiometry Moles

Name: _________________________ Class: _________________________ Date: _________________________ Score: ________ / 50

Duration: 60 minutes Total Marks: 50

Instructions:

Answer all questions in the spaces provided.
Show all working clearly for calculation questions.
The use of a data booklet and a non-programmable calculator is permitted.
Marks are awarded for correct working, not just the final answer.

Section A: Multiple Choice (Questions 1–5)

Each question is worth 1 mark. Circle the correct option.

1. A sample of hydrated sodium carbonate, $\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}$ , has a molar mass of 286 g mol⁻¹. What is the value of $x$ ? (Relative atomic masses: H = 1.0, C = 12.0, O = 16.0, Na = 23.0)

A. 5 B. 8 C. 10 D. 12

Answer space:

2. What is the total number of atoms present in 0.25 mol of ammonium sulfate, $(\text{NH}_4)_2\text{SO}_4$ ? (Avogadro constant, $N_A = 6.02 \times 10^{23}$ mol⁻¹)

A. $1.51 \times 10^{23}$ B. $4.52 \times 10^{23}$ C. $1.20 \times 10^{24}$ D. $1.81 \times 10^{24}$

Answer space:

3. A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Its relative molecular mass is 180. What is its molecular formula? (Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)

A. $\text{C}_3\text{H}_6\text{O}_3$ B. $\text{C}_4\text{H}_8\text{O}_4$ C. $\text{C}_5\text{H}_{10}\text{O}_5$ D. $\text{C}_6\text{H}_{12}\text{O}_6$

Answer space:

4. Which volume of 0.200 mol dm⁻³ sulfuric acid is required to neutralise 25.0 cm³ of 0.400 mol dm⁻³ potassium hydroxide solution?

A. 12.5 cm³ B. 25.0 cm³ C. 50.0 cm³ D. 100.0 cm³

Answer space:

5. A 0.500 g sample of an organic liquid was vaporised and found to occupy 160 cm³ at 100 °C and 100 kPa. What is the relative molecular mass of the liquid? (Gas constant, $R = 8.31$ J K⁻¹ mol⁻¹)

A. 60.0 B. 72.0 C. 84.0 D. 96.0

Answer space:

Section B: Short Answer Questions (Questions 6–10)

Answer all questions in the spaces provided. Marks are shown in brackets.

6. A student prepares a standard solution of oxalic acid, $\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}$ .

(a) Calculate the mass of hydrated oxalic acid needed to prepare 250.0 cm³ of a 0.0500 mol dm⁻³ solution. (Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)

[2]

Answer space:

(b) The student then titrates 25.0 cm³ of this oxalic acid solution against a solution of sodium hydroxide. The equation for the reaction is:

$\text{H}_2\text{C}_2\text{O}_4(\text{aq}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{C}_2\text{O}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})$

If 20.0 cm³ of NaOH solution is required for complete neutralisation, calculate the concentration of the NaOH solution.

[2]

Answer space:

7. A 1.20 g sample of a metal carbonate, $\text{MCO}_3$ , is heated strongly and decomposes completely according to the equation:

$\text{MCO}_3(\text{s}) \rightarrow \text{MO}(\text{s}) + \text{CO}_2(\text{g})$

The carbon dioxide produced occupies 268 cm³ at room temperature and pressure (r.t.p.), where the molar volume is 24.0 dm³ mol⁻¹.

(a) Calculate the amount, in mol, of carbon dioxide produced.

[1]

Answer space:

(b) Determine the relative atomic mass of metal M and identify the metal. (Relative atomic masses: C = 12.0, O = 16.0)

[3]

Answer space:

8. A 0.150 g sample of an organic compound containing only carbon, hydrogen, and oxygen is burned completely in excess oxygen. The products are 0.220 g of carbon dioxide and 0.090 g of water.

(a) Calculate the mass of carbon and hydrogen in the original sample.

[2]

Answer space:

(b) Determine the empirical formula of the compound. (Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)

[2]

Answer space:

9. A solution contains a mixture of sodium chloride and potassium chloride. A 1.000 g sample of the mixture is dissolved in water and treated with excess silver nitrate solution. The precipitate of silver chloride formed is filtered, dried, and weighed. The mass of silver chloride is 2.150 g.

(a) Write the balanced ionic equation for the precipitation reaction.

[1]

Answer space:

(b) Calculate the mass of chloride ions in the original sample. (Relative atomic masses: Cl = 35.5, Ag = 107.9)

[2]

Answer space:

[1]

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10. A student investigates the reaction between magnesium and hydrochloric acid:

$\text{Mg}(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MgCl}_2(\text{aq}) + \text{H}_2(\text{g})$

A 0.240 g sample of magnesium ribbon is added to 50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid.

(a) Determine the limiting reactant. Show your working. (Relative atomic mass: Mg = 24.3)

[2]

Answer space:

(b) Calculate the maximum volume of hydrogen gas produced at room temperature and pressure (r.t.p.), where the molar volume is 24.0 dm³ mol⁻¹.

[2]

Answer space:

Section C: Extended Response Questions (Questions 11–20)

Answer all questions in the spaces provided. Marks are shown in brackets.

11. A 0.500 g sample of a metal chloride, $\text{MCl}_2$ , is dissolved in water and treated with excess silver nitrate solution. The precipitate of silver chloride has a mass of 1.435 g.

(a) Calculate the amount, in mol, of silver chloride formed. (Relative atomic masses: Ag = 107.9, Cl = 35.5)

[1]

Answer space:

(b) Determine the relative atomic mass of metal M and identify the metal.

[3]

Answer space:

12. A student prepares a solution of potassium permanganate by dissolving 1.580 g of $\text{KMnO}_4$ in water and making the solution up to 250.0 cm³ in a volumetric flask.

(a) Calculate the concentration of the potassium permanganate solution in mol dm⁻³. (Relative atomic masses: K = 39.1, Mn = 54.9, O = 16.0)

[2]

Answer space:

(b) In a titration, 25.0 cm³ of this solution reacts exactly with 20.0 cm³ of iron(II) sulfate solution. The reaction is:

$\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l}) + 5\text{Fe}^{3+}(\text{aq})$

Calculate the concentration of the iron(II) sulfate solution.

[3]

Answer space:

13. A hydrocarbon X contains 85.7% carbon by mass. Its relative molecular mass is 84.

(a) Determine the empirical formula of X. (Relative atomic mass: H = 1.0, C = 12.0)

[2]

Answer space:

(b) Determine the molecular formula of X.

[1]

Answer space:

(c) X reacts with bromine water, which is decolourised. Suggest a possible structure for X and draw its displayed formula.

[2]

Answer space:

14. A 0.200 g sample of a metal, M, reacts completely with 50.0 cm³ of 0.500 mol dm⁻³ hydrochloric acid. The equation for the reaction is:

$\text{M}(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MCl}_2(\text{aq}) + \text{H}_2(\text{g})$

(a) Calculate the amount, in mol, of hydrochloric acid used.

[1]

Answer space:

(b) Determine the relative atomic mass of metal M and identify the metal.

[3]

Answer space:

15. A fertiliser contains ammonium sulfate, $(\text{NH}_4)_2\text{SO}_4$ , as its only nitrogen-containing component. A 2.50 g sample of the fertiliser is analysed by boiling with excess sodium hydroxide solution. The ammonia produced is absorbed in 50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid. The excess acid requires 25.0 cm³ of 0.500 mol dm⁻³ sodium hydroxide solution for neutralisation.

(a) Write the balanced equation for the reaction between ammonium sulfate and sodium hydroxide.

[1]

Answer space:

(b) Calculate the amount, in mol, of ammonia produced.

[3]

Answer space:

(c) Calculate the percentage by mass of nitrogen in the fertiliser. (Relative atomic masses: H = 1.0, N = 14.0, O = 16.0, S = 32.1)

[2]

Answer space:

16. A sample of hydrated copper(II) sulfate, $\text{CuSO}_4 \cdot x\text{H}_2\text{O}$ , is analysed by heating. A 2.50 g sample is heated until constant mass, leaving 1.60 g of anhydrous copper(II) sulfate.

(a) Calculate the mass of water lost.

[1]

Answer space:

(b) Determine the value of $x$ . (Relative atomic masses: H = 1.0, O = 16.0, S = 32.1, Cu = 63.5)

[3]

Answer space:

17. A student carries out an experiment to determine the number of moles of water of crystallisation in a sample of hydrated sodium carbonate, $\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}$ . The following results are obtained:

Mass of empty crucible: 12.50 g
Mass of crucible + hydrated salt: 16.30 g
Mass of crucible + anhydrous salt after heating: 14.30 g

(a) Calculate the mass of hydrated salt used and the mass of water lost.

[2]

Answer space:

(b) Determine the value of $x$ . (Relative atomic masses: H = 1.0, C = 12.0, O = 16.0, Na = 23.0)

[3]

Answer space:

18. A 0.100 g sample of a gaseous hydrocarbon occupies 80.0 cm³ at 27 °C and 100 kPa.

(a) Calculate the amount, in mol, of the hydrocarbon. (Gas constant, $R = 8.31$ J K⁻¹ mol⁻¹)

[2]

Answer space:

(b) The hydrocarbon contains 85.7% carbon by mass. Determine its molecular formula. (Relative atomic masses: H = 1.0, C = 12.0)

[3]

Answer space:

19. A 1.00 g sample of a mixture of calcium carbonate and calcium oxide is heated strongly. The mass of the residue after heating is 0.78 g.

(a) Write the equation for the thermal decomposition of calcium carbonate.

[1]

Answer space:

(b) Calculate the mass of calcium carbonate in the original mixture. (Relative atomic masses: C = 12.0, O = 16.0, Ca = 40.1)

[3]

Answer space:

20. A student prepares a standard solution of sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O}$ .

(a) Calculate the mass of hydrated sodium thiosulfate needed to prepare 500.0 cm³ of a 0.100 mol dm⁻³ solution. (Relative atomic masses: H = 1.0, O = 16.0, Na = 23.0, S = 32.1)

[2]

Answer space:

(b) The student uses this solution to standardise a solution of iodine. In the titration, 25.0 cm³ of the sodium thiosulfate solution reacts with 20.0 cm³ of iodine solution. The reaction is:

$2\text{S}_2\text{O}_3^{2-}(\text{aq}) + \text{I}_2(\text{aq}) \rightarrow \text{S}_4\text{O}_6^{2-}(\text{aq}) + 2\text{I}^-(\text{aq})$

Calculate the concentration of the iodine solution.

[3]

Answer space:

END OF QUIZ

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Answers

A-Level Chemistry H3 Quiz - Stoichiometry Moles: Answer Key

Total Marks: 50

Section A: Multiple Choice (Questions 1–5)

1. C. 10

Molar mass of $\text{Na}_2\text{CO}_3$ = (2 × 23.0) + 12.0 + (3 × 16.0) = 106 g mol⁻¹
Molar mass of $\text{H}_2\text{O}$ = 18.0 g mol⁻¹
Molar mass of hydrate = 106 + 18 $x$ = 286
$x$ = (286 − 106) / 18 = 180 / 18 = 10

Marking note: 1 mark for correct option.

2. D. $1.81 \times 10^{24}$

1 formula unit of $(\text{NH}_4)_2\text{SO}_4$ contains 2(N) + 8(H) + 1(S) + 4(O) = 15 atoms
Total atoms = 0.25 × 15 × $N_A$ = 0.25 × 15 × $6.02 \times 10^{23}$ = $2.26 \times 10^{24}$ atoms

Marking note: 1 mark for correct option.

3. D. $\text{C}_6\text{H}_{12}\text{O}_6$

Assume 100 g: C = 40.0 g, H = 6.7 g, O = 53.3 g
Moles: C = 40.0/12.0 = 3.33; H = 6.7/1.0 = 6.7; O = 53.3/16.0 = 3.33
Ratio: C:H:O = 3.33:6.7:3.33 = 1:2:1
Empirical formula = $\text{CH}_2\text{O}$ , empirical mass = 30
$n$ = 180/30 = 6
Molecular formula = $\text{C}_6\text{H}_{12}\text{O}_6$

Marking note: 1 mark for correct option.

4. B. 25.0 cm³

Moles of KOH = 0.400 × 25.0/1000 = 0.0100 mol
From equation: $\text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}$
Moles of $\text{H}_2\text{SO}_4$ = 0.0100/2 = 0.00500 mol
Volume = 0.00500 / 0.200 = 0.0250 dm³ = 25.0 cm³

Marking note: 1 mark for correct option.

5. D. 96.0

$pV = nRT$
$n = pV/RT = (100 \times 10^3 \times 160 \times 10^{-6}) / (8.31 \times 373)$
$n = 16.0 / 3100 = 0.00516$ mol
$M = m/n = 0.500 / 0.00516 = 96.9$ g mol⁻¹ ≈ 96.0

Marking note: 1 mark for correct option. Accept 96–97.

Section B: Short Answer Questions (Questions 6–10)

6. (a)

Moles of oxalic acid = 0.0500 × 250.0/1000 = 0.0125 mol
Molar mass of $\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}$ = (2×1.0) + (2×12.0) + (4×16.0) + (4×1.0) + (2×16.0) = 126 g mol⁻¹
Mass = 0.0125 × 126 = 1.575 g ≈ 1.58 g

Marking note: 1 mark for moles, 1 mark for mass. Accept 1.58 g.

(b)

Moles of oxalic acid in 25.0 cm³ = 0.0500 × 25.0/1000 = 0.00125 mol
From equation: 1 mol oxalic acid : 2 mol NaOH
Moles of NaOH = 2 × 0.00125 = 0.00250 mol
Concentration = 0.00250 / (20.0/1000) = 0.125 mol dm⁻³

Marking note: 1 mark for moles of NaOH, 1 mark for concentration.

7. (a)

Moles of $\text{CO}_2$ = 268/24000 = 0.01117 mol ≈ 0.0112 mol

Marking note: 1 mark.

(b)

From equation: 1 mol $\text{MCO}_3$ : 1 mol $\text{CO}_2$
Moles of $\text{MCO}_3$ = 0.01117 mol
Molar mass of $\text{MCO}_3$ = 1.20/0.01117 = 107.4 g mol⁻¹
Molar mass of $\text{CO}_3^{2-}$ = 12.0 + (3×16.0) = 60.0 g mol⁻¹
$A_r(\text{M})$ = 107.4 − 60.0 = 47.4 ≈ 47.9 → Ti (titanium)

Marking note: 1 mark for moles of carbonate, 1 mark for molar mass, 1 mark for identification. Accept Ti.

8. (a)

Mass of C = 0.220 × (12.0/44.0) = 0.0600 g
Mass of H = 0.090 × (2.0/18.0) = 0.0100 g

Marking note: 1 mark each.

(b)

Mass of O = 0.150 − 0.0600 − 0.0100 = 0.0800 g
Moles: C = 0.0600/12.0 = 0.00500; H = 0.0100/1.0 = 0.0100; O = 0.0800/16.0 = 0.00500
Ratio: C:H:O = 0.00500:0.0100:0.00500 = 1:2:1
Empirical formula = $\text{CH}_2\text{O}$

Marking note: 1 mark for mass of O, 1 mark for empirical formula.

9. (a)

$\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})$

Marking note: 1 mark. State symbols required.

(b)

Moles of AgCl = 2.150 / (107.9 + 35.5) = 2.150 / 143.4 = 0.0150 mol
From equation: 1 mol AgCl : 1 mol Cl⁻
Mass of Cl⁻ = 0.0150 × 35.5 = 0.5325 g ≈ 0.533 g

Marking note: 1 mark for moles, 1 mark for mass.

(c)

Percentage = (0.5325/1.000) × 100 = 53.3%

Marking note: 1 mark.

10. (a)

Moles of Mg = 0.240/24.3 = 0.00988 mol
Moles of HCl = 1.00 × 50.0/1000 = 0.0500 mol
From equation: 1 mol Mg : 2 mol HCl
Moles of HCl needed = 2 × 0.00988 = 0.0198 mol
Since 0.0500 mol > 0.0198 mol, HCl is in excess, so Mg is the limiting reactant.

Marking note: 1 mark for moles of both reactants, 1 mark for identifying limiting reactant.

(b)

Moles of $\text{H}_2$ = moles of Mg = 0.00988 mol
Volume = 0.00988 × 24.0 = 0.237 dm³ = 237 cm³

Marking note: 1 mark for moles of H₂, 1 mark for volume.

Section C: Extended Response Questions (Questions 11–20)

11. (a)

Moles of AgCl = 1.435 / (107.9 + 35.5) = 1.435 / 143.4 = 0.0100 mol

Marking note: 1 mark.

(b)

From equation: $\text{MCl}_2 + 2\text{AgNO}_3 \rightarrow 2\text{AgCl} + \text{M(NO}_3)_2$
Moles of $\text{MCl}_2$ = 0.0100/2 = 0.00500 mol
Molar mass of $\text{MCl}_2$ = 0.500/0.00500 = 100 g mol⁻¹
$A_r(\text{M})$ = 100 − (2 × 35.5) = 100 − 71.0 = 29.0 → No common metal at 29.0. Check: If M = Cu (63.5), $\text{CuCl}_2$ = 134.5. If M = Mg (24.3), $\text{MgCl}_2$ = 95.3. Closest is Mg (24.3), giving molar mass 95.3, not 100. Recalculate: If M = Fe (55.8), $\text{FeCl}_2$ = 126.8. If M = Ni (58.7), $\text{NiCl}_2$ = 129.7. If M = Zn (65.4), $\text{ZnCl}_2$ = 136.4. The value 100 g mol⁻¹ suggests M = 29.0, which is not a common metal. However, if the precipitate mass were different, the answer would change. Given the data, M ≈ 29.0, which is closest to Mg (24.3) but not exact. Accept Mg as the closest common metal, noting the discrepancy.

Marking note: 1 mark for moles of MCl₂, 1 mark for molar mass, 1 mark for identification. Accept Mg with note.

12. (a)

Molar mass of $\text{KMnO}_4$ = 39.1 + 54.9 + (4×16.0) = 158.0 g mol⁻¹
Moles = 1.580/158.0 = 0.0100 mol
Concentration = 0.0100 / (250.0/1000) = 0.0400 mol dm⁻³

Marking note: 1 mark for moles, 1 mark for concentration.

(b)

Moles of $\text{MnO}_4^-$ in 25.0 cm³ = 0.0400 × 25.0/1000 = 0.00100 mol
From equation: 1 mol $\text{MnO}_4^-$ : 5 mol $\text{Fe}^{2+}$
Moles of $\text{Fe}^{2+}$ = 5 × 0.00100 = 0.00500 mol
Concentration = 0.00500 / (20.0/1000) = 0.250 mol dm⁻³

Marking note: 1 mark for moles of permanganate, 1 mark for moles of Fe²⁺, 1 mark for concentration.

13. (a)

Assume 100 g: C = 85.7 g, H = 14.3 g
Moles: C = 85.7/12.0 = 7.14; H = 14.3/1.0 = 14.3
Ratio: C:H = 7.14:14.3 = 1:2
Empirical formula = $\text{CH}_2$

Marking note: 1 mark for moles, 1 mark for ratio.

(b)

Empirical mass = 14.0
$n$ = 84/14 = 6
Molecular formula = $\text{C}_6\text{H}_{12}$

Marking note: 1 mark.

(c)

X is an alkene (decolourises bromine water). Possible structure: hex-1-ene, $\text{CH}_2=\text{CHCH}_2\text{CH}_2\text{CH}_2\text{CH}_3$ , or cyclohexane (but cyclohexane does not decolourise bromine water readily). Since it decolourises bromine water, it must be an alkene.
Displayed formula: draw hex-1-ene with all bonds shown.

Marking note: 1 mark for identifying as alkene, 1 mark for correct displayed formula. Accept any valid alkene isomer of $\text{C}_6\text{H}_{12}$ .

14. (a)

Moles of HCl = 0.500 × 50.0/1000 = 0.0250 mol

Marking note: 1 mark.

(b)

From equation: 1 mol M : 2 mol HCl
Moles of M = 0.0250/2 = 0.0125 mol
$A_r(\text{M})$ = 0.200/0.0125 = 16.0 → Not a metal. Recheck: If M = Mg (24.3), moles of Mg = 0.200/24.3 = 0.00823 mol, requiring 0.0165 mol HCl. But we have 0.0250 mol HCl, so HCl is in excess. The question states M reacts completely, so M is limiting. Moles of M = 0.0125 mol, $A_r$ = 16.0, which is oxygen, not a metal. This suggests the question data is inconsistent. Correct approach: If M reacts completely, moles of M = moles of HCl/2 = 0.0125 mol, $A_r$ = 16.0. This is not a metal. Flag as data error; accept any reasonable metal if student identifies inconsistency.

Marking note: 1 mark for moles of M, 1 mark for $A_r$ , 1 mark for identification. Accept Mg if student notes HCl excess.

15. (a)

$(\text{NH}_4)_2\text{SO}_4(\text{s}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{SO}_4(\text{aq}) + 2\text{NH}_3(\text{g}) + 2\text{H}_2\text{O}(\text{l})$

Marking note: 1 mark. State symbols required.

(b)

Moles of HCl initially = 1.00 × 50.0/1000 = 0.0500 mol
Moles of NaOH for back-titration = 0.500 × 25.0/1000 = 0.0125 mol
Moles of HCl in excess = 0.0125 mol (since NaOH : HCl = 1:1)
Moles of HCl reacted with NH₃ = 0.0500 − 0.0125 = 0.0375 mol
From equation: $\text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4\text{Cl}$ , 1:1 ratio
Moles of NH₃ = 0.0375 mol

Marking note: 1 mark for initial HCl, 1 mark for excess HCl, 1 mark for moles of NH₃.

(c)

From equation in (a): 1 mol $(\text{NH}_4)_2\text{SO}_4$ : 2 mol NH₃
Moles of $(\text{NH}_4)_2\text{SO}_4$ = 0.0375/2 = 0.01875 mol
Molar mass of $(\text{NH}_4)_2\text{SO}_4$ = (2×14.0) + (8×1.0) + 32.1 + (4×16.0) = 132.1 g mol⁻¹
Mass of $(\text{NH}_4)_2\text{SO}_4$ = 0.01875 × 132.1 = 2.477 g
Mass of N = 0.01875 × 2 × 14.0 = 0.525 g
Percentage of N = (0.525/2.50) × 100 = 21.0%

Marking note: 1 mark for mass of ammonium sulfate, 1 mark for percentage.

16. (a)

Mass of water = 2.50 − 1.60 = 0.90 g

Marking note: 1 mark.

(b)

Moles of $\text{CuSO}_4$ = 1.60 / (63.5 + 32.1 + 4×16.0) = 1.60 / 159.6 = 0.01003 mol
Moles of $\text{H}_2\text{O}$ = 0.90 / 18.0 = 0.0500 mol
Ratio: $\text{CuSO}_4$ : $\text{H}_2\text{O}$ = 0.01003 : 0.0500 = 1 : 4.99 ≈ 1 : 5
$x$ = 5

Marking note: 1 mark for moles of CuSO₄, 1 mark for moles of water, 1 mark for ratio.

17. (a)

Mass of hydrated salt = 16.30 − 12.50 = 3.80 g
Mass of anhydrous salt = 14.30 − 12.50 = 1.80 g
Mass of water lost = 3.80 − 1.80 = 2.00 g

Marking note: 1 mark for mass of hydrated salt, 1 mark for mass of water.

(b)

Moles of $\text{Na}_2\text{CO}_3$ = 1.80 / 106.0 = 0.01698 mol
Moles of $\text{H}_2\text{O}$ = 2.00 / 18.0 = 0.1111 mol
Ratio: 0.01698 : 0.1111 = 1 : 6.54 ≈ 1 : 6.5. This is not a whole number. Recheck: If the mass of anhydrous salt is 1.80 g, moles = 1.80/106 = 0.01698. Water = 2.00/18 = 0.1111. Ratio = 6.54. This suggests $x$ ≈ 6.5, which is unusual. Accept $x$ = 6 or 7 with note on experimental error.

Marking note: 1 mark for moles of Na₂CO₃, 1 mark for moles of water, 1 mark for ratio. Accept 6 or 7.

18. (a)

$pV = nRT$
$n = (100 \times 10^3 \times 80.0 \times 10^{-6}) / (8.31 \times 300)$
$n = 8.00 / 2493 = 0.00321$ mol

Marking note: 1 mark for substitution, 1 mark for answer.

(b)

Molar mass = 0.100 / 0.00321 = 31.2 g mol⁻¹
Assume 100 g: C = 85.7 g, H = 14.3 g
Moles: C = 7.14, H = 14.3
Ratio: 1:2 → empirical formula = $\text{CH}_2$ , empirical mass = 14.0
$n$ = 31.2/14.0 = 2.23 ≈ 2
Molecular formula = $\text{C}_2\text{H}_4$

Marking note: 1 mark for molar mass, 1 mark for empirical formula, 1 mark for molecular formula.

19. (a)

$\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})$

Marking note: 1 mark. State symbols required.

(b)

Mass loss = 1.00 − 0.78 = 0.22 g (this is the mass of $\text{CO}_2$ )
Moles of $\text{CO}_2$ = 0.22 / 44.0 = 0.00500 mol
From equation: 1 mol $\text{CaCO}_3$ : 1 mol $\text{CO}_2$
Moles of $\text{CaCO}_3$ = 0.00500 mol
Molar mass of $\text{CaCO}_3$ = 40.1 + 12.0 + (3×16.0) = 100.1 g mol⁻¹
Mass of $\text{CaCO}_3$ = 0.00500 × 100.1 = 0.5005 g ≈ 0.50 g

Marking note: 1 mark for mass of CO₂, 1 mark for moles, 1 mark for mass of CaCO₃.

20. (a)

Molar mass of $\text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O}$ = (2×23.0) + (2×32.1) + (3×16.0) + (10×1.0) + (5×16.0) = 248.2 g mol⁻¹
Moles needed = 0.100 × 500.0/1000 = 0.0500 mol
Mass = 0.0500 × 248.2 = 12.41 g

Marking note: 1 mark for molar mass, 1 mark for mass.

(b)

Moles of $\text{S}_2\text{O}_3^{2-}$ in 25.0 cm³ = 0.100 × 25.0/1000 = 0.00250 mol
From equation: 2 mol $\text{S}_2\text{O}_3^{2-}$ : 1 mol $\text{I}_2$
Moles of $\text{I}_2$ = 0.00250/2 = 0.00125 mol
Concentration = 0.00125 / (20.0/1000) = 0.0625 mol dm⁻³

Marking note: 1 mark for moles of thiosulfate, 1 mark for moles of iodine, 1 mark for concentration.

END OF ANSWER KEY