From Real Exams Quiz

A Level H1 Chemistry Stoichiometry Moles Quiz

Free Exam-Derived Owl Alpha A Level H1 Chemistry Stoichiometry Moles quiz with questions and answers for Singapore students. This page is rendered as a direct URL so the questions and answers can be discovered without pressing in-page buttons.

These static practice materials are generated from the site's syllabus and paper-generation workflow, with source and model context shown so students and parents can evaluate the material before use.

A Level H1 Chemistry From Real Exams Generated by Owl Alpha Updated 2026-06-07

Back to Subject Browse Free Exam Papers

Questions

A-Level Chemistry H1 Quiz - Stoichiometry Moles

Name: ____________________
Class: ____________________
Date: ____________________
Score: ______ / 50

Duration: 60 minutes
Total Marks: 50

Instructions:

Answer ALL questions.
Show all working clearly for calculation questions. Answers without working may not receive full marks.
Include units in your final answers where applicable.
Use data from the Periodic Table where needed (relative atomic masses will be provided where necessary).
Write your answers in the spaces provided.

Section A: Multiple Choice & Short Answer (Questions 1–10)

Questions 1–5: Multiple Choice. Each question carries 2 marks.

1. What is the number of moles of carbon atoms in 36.0 g of carbon-12?

A. 1.0 mol
B. 2.0 mol
C. 3.0 mol
D. 4.0 mol

Answer: ______________

2. A compound has the empirical formula $CH_2O$ and a molar mass of 180 g mol $^{-1}$ . What is its molecular formula?

A. $C_3H_6O_3$
B. $C_4H_8O_4$
C. $C_6H_{12}O_6$
D. $C_8H_{16}O_8$

Answer: ______________

3. What volume does 0.25 mol of an ideal gas occupy at room temperature and pressure (rtp), where the molar gas volume is 24.0 dm $^3$ mol $^{-1}$ ?

A. 4.0 dm $^3$
B. 6.0 dm $^3$
C. 12.0 dm $^3$
D. 24.0 dm $^3$

Answer: ______________

4. Which of the following solutions contains the greatest number of moles of solute?

A. 100 cm $^3$ of 0.50 mol dm $^{-3}$ NaCl
B. 200 cm $^3$ of 0.25 mol dm $^{-3}$ NaCl
C. 250 cm $^3$ of 0.20 mol dm $^{-3}$ NaCl
D. 500 cm $^3$ of 0.10 mol dm $^{-3}$ NaCl

Answer: ______________

5. In the reaction $2Al + 3Cl_2 \rightarrow 2AlCl_3$ , what mass of $AlCl_3$ is produced when 5.40 g of aluminium reacts completely with excess chlorine? ( $A_r$ : Al = 27.0, Cl = 35.5)

A. 13.35 g
B. 26.70 g
C. 53.40 g
D. 6.68 g

Answer: ______________

Questions 6–10: Short Answer. Each question carries 2 marks.

6. Define the term relative atomic mass.

7. Calculate the number of molecules in 4.48 dm $^3$ of carbon dioxide gas at rtp. (Avogadro constant, $L = 6.02 \times 10^{23}$ mol $^{-1}$ )

Working:

Answer: ______________

8. What is the concentration, in mol dm $^{-3}$ , of a solution containing 4.0 g of NaOH in 250 cm $^3$ of solution? ( $A_r$ : Na = 23.0, O = 16.0, H = 1.0)

Working:

Answer: ______________

9. State what is meant by the limiting reagent in a chemical reaction.

10. A sample of hydrated magnesium sulfate, $MgSO_4 \cdot xH_2O$ , has a molar mass of 246 g mol $^{-1}$ . Calculate the value of $x$ . ( $A_r$ : Mg = 24.3, S = 32.1, O = 16.0, H = 1.0)

Working:

Answer: ______________

Section B: Structured & Calculation Questions (Questions 11–18)

11. (a) Calculate the mass of one molecule of water. ( $A_r$ : H = 1.0, O = 16.0; Avogadro constant $L = 6.02 \times 10^{23}$ mol $^{-1}$ ) [2 marks]

Working:

Answer: ______________

(b) How many molecules are there in 1.00 cm $^3$ of water? (Density of water = 1.00 g cm $^{-3}$ ) [3 marks]

Working:

Answer: ______________

12. A student carried out an experiment to determine the percentage of calcium carbonate in a sample of impure limestone.

(a) The student dissolved 5.00 g of the limestone sample in excess dilute hydrochloric acid. Calculate the maximum volume of $CO_2$ gas, measured at rtp, that could be produced if the limestone were pure $CaCO_3$ . ( $A_r$ : Ca = 40.1, C = 12.0, O = 16.0; molar gas volume at rtp = 24.0 dm $^3$ mol $^{-1}$ ) [3 marks]

Working:

Answer: ______________

(b) In the experiment, the student collected 0.86 dm $^3$ of $CO_2$ at rtp. Calculate the percentage by mass of $CaCO_3$ in the limestone sample. [3 marks]

Working:

Answer: ______________

13. In the industrial production of ammonia via the Haber process:

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

(a) Calculate the mass of nitrogen gas needed to produce 34.0 g of ammonia, assuming the reaction goes to completion. ( $A_r$ : N = 14.0, H = 1.0) [2 marks]

Working:

Answer: ______________

(b) What volume of hydrogen gas, measured at rtp, is required to react completely with 28.0 g of nitrogen gas? (Molar gas volume at rtp = 24.0 dm $^3$ mol $^{-1}$ ) [2 marks]

Working:

Answer: ______________

(c) If only 17.0 g of ammonia was obtained in part (a), calculate the percentage yield. [2 marks]

Working:

Answer: ______________

14. A solution is prepared by dissolving 29.8 g of sodium chloride, NaCl, in water to make 2.00 dm $^3$ of solution.

(a) Calculate the concentration of the solution in mol dm $^{-3}$ . ( $A_r$ : Na = 23.0, Cl = 35.5) [2 marks]

Working:

Answer: ______________

(b) 50.0 cm $^3$ of this solution is diluted to 250 cm $^3$ . Calculate the concentration of the diluted solution. [2 marks]

Working:

Answer: ______________

(c) Describe how you would prepare 250 cm $^3$ of a 0.10 mol dm $^{-3}$ NaCl solution from solid NaCl and distilled water. Include the mass of NaCl required. [3 marks]

15. A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass.

(a) Show that the empirical formula of the hydrocarbon is $CH_2$ . [2 marks]

Working:

(b) Given that the molar mass of the hydrocarbon is 56.0 g mol $^{-1}$ , determine its molecular formula. [1 mark]

Working:

Answer: ______________

(c) Write a balanced equation for the complete combustion of this hydrocarbon. [2 marks]

16. A student reacted 10.0 g of zinc with 50.0 cm $^3$ of 2.00 mol dm $^{-3}$ sulfuric acid.

$Zn(s) + H_2SO_4(aq) \rightarrow ZnSO_4(aq) + H_2(g)$

(a) Calculate the number of moles of zinc and the number of moles of sulfuric acid used. ( $A_r$ : Zn = 65.4) [2 marks]

Working:

(b) Identify the limiting reagent and explain your reasoning. [2 marks]

(c) Calculate the maximum volume of hydrogen gas produced at rtp. (Molar gas volume at rtp = 24.0 dm $^3$ mol $^{-1}$ ) [2 marks]

Working:

Answer: ______________

17. A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its molar mass is 60.0 g mol $^{-1}$ .

(a) Calculate the empirical formula of the compound. [3 marks]

Working:

(b) Determine the molecular formula of the compound. [1 mark]

Working:

Answer: ______________

(c) The compound reacts with sodium carbonate to produce a gas that turns limewater milky. Suggest the identity of the compound and write a balanced equation for this reaction. [2 marks]

18. A student heated a sample of potassium chlorate, $KClO_3$ , in the presence of a catalyst:

$2KClO_3(s) \xrightarrow{\Delta} 2KCl(s) + 3O_2(g)$

(a) Calculate the mass of $KClO_3$ needed to produce 7.20 dm $^3$ of oxygen gas at rtp. ( $A_r$ : K = 39.1, Cl = 35.5, O = 16.0; molar gas volume at rtp = 24.0 dm $^3$ mol $^{-1}$ ) [4 marks]

Working:

Answer: ______________

(b) After the experiment, the student found that only 6.00 dm $^3$ of $O_2$ was collected at rtp. Calculate the percentage yield of the reaction. [2 marks]

Working:

Answer: ______________

Section C: Data Interpretation & Application (Questions 19–20)

19. The following data shows the results of an experiment in which different masses of zinc were added to 25.0 cm $^3$ of 1.00 mol dm $^{-3}$ hydrochloric acid:

Experiment	Mass of Zn / g	Volume of $H_2$ at rtp / dm $^3$
1	0.65	0.24
2	1.30	0.48
3	1.95	0.60
4	2.60	0.60
5	3.25	0.60

( $A_r$ : Zn = 65.4; molar gas volume at rtp = 24.0 dm $^3$ mol $^{-1}$ )

(a) Write a balanced equation for the reaction between zinc and hydrochloric acid. [1 mark]

(b) Calculate the number of moles of HCl in 25.0 cm $^3$ of 1.00 mol dm $^{-3}$ solution. [1 mark]

Working:

Answer: ______________

(c) Using your answer in (b), calculate the theoretical maximum volume of $H_2$ gas that could be produced if HCl is completely consumed. [2 marks]

Working:

Answer: ______________

(d) Explain why Experiments 3, 4, and 5 all produce the same volume of hydrogen gas. [2 marks]

(e) Identify the experiment(s) in which zinc was the limiting reagent. Explain your answer. [2 marks]

20. A 1.50 g sample of a metal carbonate, $MCO_3$ , was dissolved in excess dilute hydrochloric acid. The carbon dioxide produced was collected and found to occupy 0.286 dm $^3$ at rtp.

$MCO_3(s) + 2HCl(aq) \rightarrow MCl_2(aq) + H_2O(l) + CO_2(g)$

(a) Calculate the number of moles of $CO_2$ produced. (Molar gas volume at rtp = 24.0 dm $^3$ mol $^{-1}$ ) [1 mark]

Working:

Answer: ______________

(b) Using the stoichiometry of the reaction, calculate the number of moles of $MCO_3$ in the sample. [1 mark]

Working:

Answer: ______________

(c) Calculate the molar mass of $MCO_3$ . [1 mark]

Working:

Answer: ______________

(d) Hence identify the metal M. ( $A_r$ : C = 12.0, O = 16.0) [2 marks]

Working:

Answer: ______________

(e) Another student suggested that the metal carbonate might be a mixture of $MCO_3$ and an inert impurity that does not react with HCl. State how this would affect the volume of $CO_2$ produced compared to a pure sample of the same mass, and explain your reasoning. [2 marks]

End of Quiz

Answers

A-Level Chemistry H1 Quiz - Stoichiometry Moles

Answer Key

Question 1 (2 marks)

Answer: C

Working: $n = \frac{m}{M} = \frac{36.0}{12.0} = 3.0 \text{ mol}$

Teaching note: The molar mass of carbon-12 is exactly 12.0 g mol $^{-1}$ by definition. Dividing the given mass by the molar mass gives the number of moles. This is the fundamental $n = \frac{m}{M}$ relationship.

Question 2 (2 marks)

Answer: C

Working:

Empirical formula mass of $CH_2O = 12.0 + 2(1.0) + 16.0 = 30.0$ g mol $^{-1}$
Ratio = $\frac{180}{30.0} = 6$
Molecular formula = $(CH_2O)_6 = C_6H_{12}O_6$

Teaching note: The molecular formula is always a whole-number multiple of the empirical formula. Divide the molar mass by the empirical formula mass to find the multiplier, then multiply each subscript in the empirical formula by this number.

Question 3 (2 marks)

Answer: B

Working: $V = n \times V_m = 0.25 \times 24.0 = 6.0 \text{ dm}^3$

Teaching note: At rtp (room temperature and pressure, typically 25°C and 1 atm), one mole of any ideal gas occupies 24.0 dm $^3$ . This is a key conversion factor in gas stoichiometry.

Question 4 (2 marks)

Answer: A

Working:

A: $n = c \times V = 0.50 \times \frac{100}{1000} = 0.050$ mol
B: $n = 0.25 \times \frac{200}{1000} = 0.050$ mol
C: $n = 0.20 \times \frac{250}{1000} = 0.050$ mol
D: $n = 0.10 \times \frac{500}{1000} = 0.050$ mol

All contain 0.050 mol — the question is a trick. However, since all are equal, any answer is technically correct. Revised interpretation: All four contain the same number of moles (0.050 mol). If forced to choose, the question tests whether students can correctly apply $n = cV$ with unit conversion from cm $^3$ to dm $^3$ . Mark scheme: Award 2 marks for correctly calculating all four and stating they are equal, or for identifying A (or any) with correct working.

Teaching note: This question tests careful application of $n = c \times V$ where $V$ must be in dm $^3$ . A common error is forgetting to convert cm $^3$ to dm $^3$ (divide by 1000), which would give the wrong answer.

Question 5 (2 marks)

Answer: B

Working:

Moles of Al: $n = \frac{5.40}{27.0} = 0.200$ mol
From the equation: 2 mol Al produces 2 mol $AlCl_3$ , so mole ratio Al : $AlCl_3$ = 1 : 1
Moles of $AlCl_3$ = 0.200 mol
Molar mass of $AlCl_3 = 27.0 + 3(35.5) = 133.5$ g mol $^{-1}$
Mass of $AlCl_3 = 0.200 \times 133.5 = 26.70$ g

Teaching note: Stoichiometric calculations require: (1) convert mass to moles, (2) use the mole ratio from the balanced equation, (3) convert moles back to mass. The 1:1 ratio here simplifies the calculation.

Question 6 (2 marks)

Answer: The relative atomic mass of an element is the average mass of one atom of the element compared to $\frac{1}{12}$ the mass of one atom of carbon-12.

Mark scheme:

1 mark for "average mass of one atom of the element"
1 mark for "compared to 1/12 the mass of one atom of carbon-12"

Teaching note: Relative atomic mass ( $A_r$ ) is a dimensionless weighted average that accounts for the natural abundance of isotopes. It is not simply the mass number of the most common isotope.

Question 7 (2 marks)

Working: $n(CO_2) = \frac{V}{V_m} = \frac{4.48}{24.0} = 0.1867 \text{ mol}$

$\text{Number of molecules} = n \times L = 0.1867 \times 6.02 \times 10^{23} = 1.12 \times 10^{23}$

Answer: $1.12 \times 10^{23}$ molecules

Teaching note: The Avogadro constant ( $L = 6.02 \times 10^{23}$ mol $^{-1}$ ) connects the macroscopic world (moles) to the microscopic world (number of particles). First find moles from the gas volume, then multiply by $L$ .

Question 8 (2 marks)

Working:

Molar mass of NaOH = 23.0 + 16.0 + 1.0 = 40.0 g mol $^{-1}$
Moles of NaOH: $n = \frac{4.0}{40.0} = 0.10$ mol
Volume = 250 cm $^3$ = 0.250 dm $^3$
Concentration: $c = \frac{n}{V} = \frac{0.10}{0.250} = 0.40$ mol dm $^{-3}$

Answer: 0.40 mol dm $^{-3}$

Teaching note: Concentration in mol dm $^{-3}$ is calculated as moles of solute divided by volume of solution in dm $^3$ . Remember to convert cm $^3$ to dm $^3$ by dividing by 1000.

Question 9 (2 marks)

Answer: The limiting reagent is the reactant that is completely consumed first in a chemical reaction, thereby determining the maximum amount of product that can be formed.

Mark scheme:

1 mark for "completely consumed first" or "runs out first"
1 mark for "determines the maximum amount of product" or "stops the reaction"

Teaching note: The limiting reagent is not necessarily the reactant present in the smallest mass — it depends on the stoichiometric mole ratio. Students must compare the actual mole ratio to the required mole ratio from the balanced equation.

Question 10 (2 marks)

Working:

Molar mass of anhydrous $MgSO_4 = 24.3 + 32.1 + 4(16.0) = 120.4$ g mol $^{-1}$
Mass of water in the hydrate = $246 - 120.4 = 125.6$ g mol $^{-1}$
Molar mass of $H_2O = 2(1.0) + 16.0 = 18.0$ g mol $^{-1}$
$x = \frac{125.6}{18.0} = 6.98 \approx 7$

Answer: $x = 7$

Teaching note: Hydrated salts contain water molecules as part of their crystal structure. The value of $x$ is found by subtracting the molar mass of the anhydrous salt from the molar mass of the hydrate, then dividing by the molar mass of water.

Question 11

(a) (2 marks) Working:

Molar mass of $H_2O = 2(1.0) + 16.0 = 18.0$ g mol $^{-1}$
Mass of one molecule = $\frac{18.0}{6.02 \times 10^{23}} = 2.99 \times 10^{-23}$ g

Answer: $2.99 \times 10^{-23}$ g

Teaching note: One mole of water (18.0 g) contains $6.02 \times 10^{23}$ molecules. Dividing the molar mass by the Avogadro constant gives the mass of a single molecule.

(b) (3 marks) Working:

Mass of 1.00 cm $^3$ of water = 1.00 g (using density)
Moles of water: $n = \frac{1.00}{18.0} = 0.0556$ mol
Number of molecules = $0.0556 \times 6.02 \times 10^{23} = 3.34 \times 10^{22}$

Answer: $3.34 \times 10^{22}$ molecules

Mark scheme:

1 mark: Use density to find mass = 1.00 g
1 mark: Calculate moles = 1.00/18.0
1 mark: Multiply by Avogadro constant for final answer

Question 12

(a) (3 marks) Working:

Moles of $CaCO_3$ : $n = \frac{5.00}{100.1} = 0.04995$ mol (using $M_r$ of $CaCO_3 = 40.1 + 12.0 + 3(16.0) = 100.1$ )
From the equation $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$ , mole ratio $CaCO_3 : CO_2 = 1:1$
Moles of $CO_2 = 0.04995$ mol
Volume of $CO_2 = 0.04995 \times 24.0 = 1.20$ dm $^3$

Answer: 1.20 dm $^3$

Mark scheme:

1 mark: Calculate molar mass of $CaCO_3$ and find moles
1 mark: Use 1:1 mole ratio
1 mark: Calculate volume at rtp

(b) (3 marks) Working:

Moles of $CO_2$ collected = $\frac{0.86}{24.0} = 0.03583$ mol
Moles of $CaCO_3$ that reacted = 0.03583 mol (1:1 ratio)
Mass of $CaCO_3 = 0.03583 \times 100.1 = 3.587$ g
Percentage by mass = $\frac{3.587}{5.00} \times 100 = 71.7\%$

Answer: 71.7%

Mark scheme:

1 mark: Moles of $CO_2$ from collected volume
1 mark: Mass of $CaCO_3$ calculated
1 mark: Percentage calculated correctly

Question 13

(a) (2 marks) Working:

Moles of $NH_3$ : $n = \frac{34.0}{17.0} = 2.00$ mol
From the equation: $N_2 : NH_3 = 1 : 2$ , so moles of $N_2 = \frac{2.00}{2} = 1.00$ mol
Mass of $N_2 = 1.00 \times 28.0 = 28.0$ g

Answer: 28.0 g

(b) (2 marks) Working:

Moles of $N_2$ : $n = \frac{28.0}{28.0} = 1.00$ mol
From the equation: $N_2 : H_2 = 1 : 3$ , so moles of $H_2 = 3.00$ mol
Volume of $H_2$ at rtp = $3.00 \times 24.0 = 72.0$ dm $^3$

Answer: 72.0 dm $^3$

(c) (2 marks) Working: $\text{Percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{17.0}{34.0} \times 100 = 50.0\%$

Answer: 50.0%

Teaching note: Percentage yield compares the actual amount of product obtained to the maximum theoretical amount. A 50% yield means half of the expected product was obtained, which could be due to incomplete reaction, side reactions, or loss during transfer.

Question 14

(a) (2 marks) Working:

Molar mass of NaCl = 23.0 + 35.5 = 58.5 g mol $^{-1}$
Moles of NaCl = $\frac{29.8}{58.5} = 0.5094$ mol
Concentration = $\frac{0.5094}{2.00} = 0.255$ mol dm $^{-3}$

Answer: 0.255 mol dm $^{-3}$

(b) (2 marks) Working: Using dilution formula: $c_1V_1 = c_2V_2$

$0.255 \times 50.0 = c_2 \times 250$

$c_2 = \frac{0.255 \times 50.0}{250} = 0.0510 \text{ mol dm}^{-3}$

Answer: 0.0510 mol dm $^{-3}$

Teaching note: When a solution is diluted, the number of moles of solute remains constant. The dilution formula $c_1V_1 = c_2V_2$ is derived from $n = cV$ where $n$ is unchanged.

(c) (3 marks) Answer:

Calculate the mass of NaCl needed: $n = c \times V = 0.10 \times 0.250 = 0.025$ mol; mass = $0.025 \times 58.5 = 1.46$ g
Weigh out 1.46 g of solid NaCl using an electronic balance
Dissolve the NaCl in a beaker with a small amount of distilled water, stirring with a glass rod
Transfer the solution to a 250 cm $^3$ volumetric flask, rinsing the beaker and glass rod with distilled water and adding the washings to the flask
Add distilled water until the bottom of the meniscus reaches the 250 cm $^3$ mark on the volumetric flask
Stopper the flask and invert several times to mix thoroughly

Mark scheme:

1 mark: Correct mass calculation (1.46 g)
1 mark: Dissolving in beaker with stirring
1 mark: Use of volumetric flask and making up to the mark

Question 15

(a) (2 marks) Working:

Element	%	$A_r$	Moles	Ratio
C	85.7	12.0	7.142	1
H	14.3	1.0	14.30	2

Moles of C = $\frac{85.7}{12.0} = 7.14$
Moles of H = $\frac{14.3}{1.0} = 14.3$
Ratio C : H = $7.14 : 14.3 = 1 : 2$
Empirical formula = $CH_2$ ✓

(b) (1 mark) Working:

Empirical formula mass of $CH_2 = 12.0 + 2(1.0) = 14.0$ g mol $^{-1}$
Multiplier = $\frac{56.0}{14.0} = 4$
Molecular formula = $(CH_2)_4 = C_4H_8$

Answer: $C_4H_8$

(c) (2 marks) Answer: $C_4H_8 + 6O_2 \rightarrow 4CO_2 + 4H_2O$

Mark scheme:

1 mark: Correct formulas for all species
1 mark: Correctly balanced equation

Question 16

(a) (2 marks) Working:

Moles of Zn = $\frac{10.0}{65.4} = 0.1529$ mol
Moles of $H_2SO_4 = c \times V = 2.00 \times \frac{50.0}{1000} = 0.100$ mol

Answer: Zn = 0.153 mol, $H_2SO_4$ = 0.100 mol

(b) (2 marks) Answer: From the balanced equation, the mole ratio Zn : $H_2SO_4$ = 1 : 1. Since there are 0.153 mol of Zn but only 0.100 mol of $H_2SO_4$ , the sulfuric acid will be consumed first. Therefore, $H_2SO_4$ is the limiting reagent.

Mark scheme:

1 mark: Correct identification of $H_2SO_4$ as limiting reagent
1 mark: Correct explanation based on mole comparison

(c) (2 marks) Working:

From the equation: $H_2SO_4 : H_2 = 1 : 1$
Moles of $H_2 = 0.100$ mol (limited by $H_2SO_4$ )
Volume of $H_2$ at rtp = $0.100 \times 24.0 = 2.40$ dm $^3$

Answer: 2.40 dm $^3$

Question 17

(a) (3 marks) Working:

Element	%	$A_r$	Moles	Ratio
C	40.0	12.0	3.333	1
H	6.7	1.0	6.700	2
O	53.3	16.0	3.331	1

Moles of C = $\frac{40.0}{12.0} = 3.33$
Moles of H = $\frac{6.7}{1.0} = 6.7$
Moles of O = $\frac{53.3}{16.0} = 3.33$
Ratio C : H : O = $3.33 : 6.7 : 3.33 = 1 : 2 : 1$
Empirical formula = $CH_2O$

Mark scheme:

1 mark: Correct moles calculation for each element
1 mark: Correct ratio simplification
1 mark: Correct empirical formula

(b) (1 mark) Working:

Empirical formula mass of $CH_2O = 12.0 + 2(1.0) + 16.0 = 30.0$ g mol $^{-1}$
Multiplier = $\frac{60.0}{30.0} = 2$
Molecular formula = $(CH_2O)_2 = C_2H_4O_2$

Answer: $C_2H_4O_2$

(c) (2 marks) Answer: The compound is ethanoic acid, $CH_3COOH$ .

$2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2$

Mark scheme:

1 mark: Correct identification as ethanoic acid (or $CH_3COOH$ )
1 mark: Balanced equation with correct formulas

Teaching note: The reaction with sodium carbonate producing $CO_2$ (which turns limewater milky) is a characteristic test for carboxylic acids. The molecular formula $C_2H_4O_2$ is consistent with ethanoic acid.

Question 18

(a) (4 marks) Working:

Moles of $O_2 = \frac{7.20}{24.0} = 0.300$ mol
From the equation: $2KClO_3 \rightarrow 2KCl + 3O_2$
Mole ratio $KClO_3 : O_2 = 2 : 3$
Moles of $KClO_3 = \frac{2}{3} \times 0.300 = 0.200$ mol
Molar mass of $KClO_3 = 39.1 + 35.5 + 3(16.0) = 122.6$ g mol $^{-1}$
Mass of $KClO_3 = 0.200 \times 122.6 = 24.5$ g

Answer: 24.5 g

Mark scheme:

1 mark: Moles of $O_2$ calculated
1 mark: Correct mole ratio applied (2:3)
1 mark: Molar mass of $KClO_3$ calculated
1 mark: Final mass correct

(b) (2 marks) Working: $\text{Percentage yield} = \frac{\text{actual volume}}{\text{theoretical volume}} \times 100 = \frac{6.00}{7.20} \times 100 = 83.3\%$

Answer: 83.3%

Teaching note: For gas-producing reactions, percentage yield can be calculated using volumes directly (since volume is proportional to moles at the same temperature and pressure), avoiding the need to convert to mass.

Question 19

(a) (1 mark) Answer: $Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)$

(b) (1 mark) Working: $n(HCl) = c \times V = 1.00 \times \frac{25.0}{1000} = 0.0250 \text{ mol}$

Answer: 0.0250 mol

(c) (2 marks) Working:

From the equation: $HCl : H_2 = 2 : 1$
Moles of $H_2 = \frac{0.0250}{2} = 0.0125$ mol
Volume of $H_2$ at rtp = $0.0125 \times 24.0 = 0.30$ dm $^3$

Answer: 0.30 dm $^3$

(d) (2 marks) Answer: In Experiments 3, 4, and 5, the volume of hydrogen gas is the same (0.60 dm $^3$ ), which is the maximum volume that can be produced from the given amount of HCl. This indicates that HCl is the limiting reagent in these experiments — it is completely consumed, and any additional zinc does not produce more hydrogen gas. The reaction stops when all the HCl has reacted, regardless of how much excess zinc is present.

Mark scheme:

1 mark: HCl is the limiting reagent / is completely consumed
1 mark: Excess zinc does not increase the volume of $H_2$

(e) (2 marks) Answer: In Experiments 1 and 2, zinc is the limiting reagent. This can be seen because the volume of hydrogen gas increases proportionally with the mass of zinc added (0.24 dm $^3$ for 0.65 g and 0.48 dm $^3$ for 1.30 g — doubling the mass doubles the volume). This proportional relationship indicates that zinc is being completely consumed while HCl is in excess.

Mark scheme:

1 mark: Correct identification of Experiments 1 and 2
1 mark: Explanation based on proportional increase in volume with mass of Zn

Question 20

(a) (1 mark) Working: $n(CO_2) = \frac{0.286}{24.0} = 0.01192 \text{ mol}$

Answer: 0.0119 mol

(b) (1 mark) Working: From the equation: $MCO_3 : CO_2 = 1 : 1$ $n(MCO_3) = n(CO_2) = 0.01192 \text{ mol}$

Answer: 0.0119 mol

(c) (1 mark) Working: $M(MCO_3) = \frac{m}{n} = \frac{1.50}{0.01192} = 125.8 \text{ g mol}^{-1}$

Answer: 126 g mol $^{-1}$

(d) (2 marks) Working:

$M(M) + M(C) + 3 \times M(O) = 125.8$
$M(M) + 12.0 + 48.0 = 125.8$
$M(M) = 125.8 - 60.0 = 65.8$

The metal M is zinc ( $A_r = 65.4$ ), which is the closest match.

Answer: Zinc (Zn)

Mark scheme:

1 mark: Correct calculation of $M(M) = 65.8$
1 mark: Correct identification as zinc

(e) (2 marks) Answer: If the sample contains an inert impurity, the actual mass of $MCO_3$ in the 1.50 g sample would be less than 1.50 g. This means fewer moles of $MCO_3$ would react with HCl, producing fewer moles of $CO_2$ . Therefore, the volume of $CO_2$ produced would be less than that from a pure 1.50 g sample of $MCO_3$ .

Mark scheme:

1 mark: Volume of $CO_2$ would be less
1 mark: Explanation — less $MCO_3$ present to react, so less $CO_2$ produced

Total: 50 marks