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A Level H1 Chemistry Stoichiometry Moles Quiz

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Questions

A-Level Chemistry H1 Quiz - Stoichiometry Moles

Name: __________________________
Class: __________________________
Date: __________________________
Score: _________ / 40

Duration: 45 minutes
Total Marks: 40
Instructions:

Answer all questions.
Show all working for calculation questions.
State units where appropriate.
Use the Data Booklet where necessary.

Section A: Multiple Choice & Short Concepts (10 Marks)

1. Which of the following contains the greatest number of atoms? [1]
A. 1.0 mol of He
B. 1.0 mol of H₂
C. 0.5 mol of CH₄
D. 0.25 mol of C₆H₁₂O₆

2. A sample of gas occupies 240 cm³ at room temperature and pressure (r.t.p.). What is the amount of gas in moles? (Molar volume of gas at r.t.p. = 24.0 dm³ mol⁻¹) [1]
A. 0.010 mol
B. 0.10 mol
C. 1.0 mol
D. 10.0 mol

3. What is the empirical formula of a compound with the molecular formula C₆H₁₂O₂? [1]
A. C₃H₆O
B. C₆H₁₂O₂
C. CH₂O
D. C₂H₄O

4. Balance the following equation and determine the coefficient of O₂:
__ C₃H₈ + __ O₂ → __ CO₂ + __ H₂O [1]
A. 3
B. 4
C. 5
D. 6

5. 20.0 cm³ of 0.100 mol dm⁻³ NaOH reacts completely with 25.0 cm³ of H₂SO₄. What is the concentration of the H₂SO₄? [1]
2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)
A. 0.040 mol dm⁻³
B. 0.080 mol dm⁻³
C. 0.100 mol dm⁻³
D. 0.200 mol dm⁻³

6. Define the term limiting reagent. [1]

7. A student dissolves 5.85 g of NaCl in water to make 500 cm³ of solution. Calculate the concentration of the solution in mol dm⁻³. (Ar: Na = 23.0, Cl = 35.5) [2]

8. Calculate the number of electrons present in 0.1 mol of OH⁻ ions. (Avogadro constant, $L$ = 6.02 × 10²³ mol⁻¹) [2]

9. Which statement about the mole is correct? [1]
A. The mass of 1 mole of any substance is the same.
B. 1 mole of any gas occupies 24.0 dm³ at all temperatures and pressures.
C. 1 mole of any substance contains the same number of particles.
D. The molar mass of an element is numerically equal to its atomic number.

10. What is the mass of 0.50 mol of calcium carbonate, CaCO₃? (Ar: Ca = 40.1, C = 12.0, O = 16.0) [1]
A. 25.0 g
B. 50.1 g
C. 100.1 g
D. 200.2 g

Section B: Structured Calculations (18 Marks)

11. Magnesium reacts with hydrochloric acid according to the equation:
Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

In an experiment, 0.12 g of magnesium is added to 50.0 cm³ of 0.50 mol dm⁻³ HCl.

(a) Calculate the amount, in moles, of magnesium used. (Ar: Mg = 24.3) [1]

(b) Calculate the amount, in moles, of HCl present in the solution. [1]

(d) Calculate the maximum volume of hydrogen gas produced at r.t.p. (Molar volume = 24.0 dm³ mol⁻¹) [2]

12. A hydrated salt of sodium carbonate has the formula Na₂CO₃·xH₂O.
2.86 g of the hydrated salt is heated strongly until all water of crystallisation is removed. The mass of the remaining anhydrous Na₂CO₃ is 1.06 g.

(a) Calculate the amount, in moles, of anhydrous Na₂CO₃ remaining. (Ar: Na = 23.0, C = 12.0, O = 16.0) [2]

(b) Calculate the mass of water lost. [1]

(d) Determine the value of x in the formula Na₂CO₃·xH₂O. [2]

13. Titanium(IV) chloride reacts with water to form titanium(IV) oxide and hydrogen chloride gas.
TiCl₄(l) + 2H₂O(l) → TiO₂(s) + 4HCl(g)

(a) Calculate the mass of TiO₂ produced when 10.0 g of TiCl₄ reacts with excess water. (Ar: Ti = 47.9, Cl = 35.5, O = 16.0) [3]

(b) If the actual yield of TiO₂ was 3.50 g, calculate the percentage yield. [2]

14. 25.0 cm³ of 0.200 mol dm⁻³ sulfuric acid, H₂SO₄, is neutralized by 20.0 cm³ of potassium hydroxide solution, KOH.
H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l)

(a) Calculate the moles of H₂SO₄ used. [1]

(b) Calculate the moles of KOH required for neutralization. [1]

15. A metal M reacts with chlorine to form a chloride with the formula MCl₂.
0.48 g of metal M reacts completely with chlorine to form 1.90 g of the chloride. (Ar: Cl = 35.5)

(a) Calculate the mass of chlorine that combined with the metal. [1]

(b) Calculate the moles of chlorine atoms in the chloride formed. [1]

Section C: Data Interpretation & Application (12 Marks)

16. A mixture of two gases, nitrogen (N₂) and oxygen (O₂), has a total mass of 10.0 g and occupies a volume of 8.00 dm³ at r.t.p.

(a) Calculate the total amount of gas molecules in the mixture in moles. [1]

(b) Let $n_{N2}$ be the moles of nitrogen and $n_{O2}$ be the moles of oxygen. Write two simultaneous equations representing the total moles and total mass of the mixture. (Ar: N = 14.0, O = 16.0) [2]

17. An organic compound Z contains carbon, hydrogen, and oxygen only.
Combustion of 0.46 g of Z produces 0.88 g of CO₂ and 0.54 g of H₂O.

(a) Calculate the mass of carbon in the original sample. [1]

(b) Calculate the mass of hydrogen in the original sample. [1]

(d) Determine the empirical formula of compound Z. [3]

18. Aluminum reacts with iron(III) oxide in the thermite reaction:
2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(l)

(a) Calculate the mass of aluminum required to react completely with 10.0 g of Fe₂O₃. (Ar: Al = 27.0, Fe = 55.8, O = 16.0) [3]

(b) If 5.0 g of Al is reacted with 10.0 g of Fe₂O₃, identify the limiting reagent. [1]

19. A solution of sodium thiosulfate, Na₂S₂O₃, is prepared by dissolving 12.4 g of the hydrated salt Na₂S₂O₃·5H₂O in water to make 250 cm³ of solution. (Ar: Na = 23.0, S = 32.1, O = 16.0, H = 1.0)

(a) Calculate the molar mass of Na₂S₂O₃·5H₂O. [1]

(b) Calculate the concentration of the sodium thiosulfate solution in mol dm⁻³. [2]

20. Gas X has the empirical formula CH₂. At r.t.p., 1.00 dm³ of gas X has a mass of 1.75 g. (Molar volume at r.t.p. = 24.0 dm³ mol⁻¹)

(a) Calculate the molar mass of gas X. [2]

(b) Determine the molecular formula of gas X. [1]

End of Quiz

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Answers

A-Level Chemistry H1 Quiz - Stoichiometry Moles (Answer Key)

1. D
Reasoning:
A: 1.0 mol He = 1.0 mol atoms.
B: 1.0 mol H₂ = 2.0 mol atoms.
C: 0.5 mol CH₄ = 0.5 × 5 = 2.5 mol atoms.
D: 0.25 mol C₆H₁₂O₆ = 0.25 × 24 = 6.0 mol atoms.
Greatest number of atoms is D.

2. A
$n = V / V_m = 240 \text{ cm}^3 / 24000 \text{ cm}^3 \text{ mol}^{-1} = 0.010 \text{ mol}$ .

3. A
Divide subscripts by greatest common divisor (2): C₆H₁₂O₂ → C₃H₆O.

4. C
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Coefficient of O₂ is 5.

5. A
Moles NaOH = $0.020 \times 0.100 = 0.002$ mol.
Ratio NaOH:H₂SO₄ is 2:1.
Moles H₂SO₄ = $0.002 / 2 = 0.001$ mol.
Conc H₂SO₄ = $0.001 / 0.025 = 0.040$ mol dm⁻³.

6.
The reactant that is completely consumed first in a chemical reaction, thereby limiting the amount of product formed. [1]

7.
Molar mass NaCl = $23.0 + 35.5 = 58.5$ g mol⁻¹.
Moles NaCl = $5.85 / 58.5 = 0.100$ mol.
Volume = $500 \text{ cm}^3 = 0.500 \text{ dm}^3$ .
Concentration = $0.100 / 0.500 = 0.200$ mol dm⁻³. [2]

8.
Electrons in one OH⁻ ion: O (8) + H (1) + charge (1) = 10 electrons.
Total moles of electrons = $0.1 \text{ mol ions} \times 10 = 1.0$ mol electrons.
Number of electrons = $1.0 \times 6.02 \times 10^{23} = 6.02 \times 10^{23}$ . [2]

9. C
A is incorrect (masses differ). B is incorrect (only at r.t.p./s.t.p.). D is incorrect (equal to relative atomic mass, not atomic number). C is the definition of the mole.

10. B
Mr CaCO₃ = $40.1 + 12.0 + (3 \times 16.0) = 100.1$ g mol⁻¹.
Mass = $0.50 \times 100.1 = 50.05$ g ≈ 50.1 g.

11.
(a) Moles Mg = $0.12 / 24.3 = 0.00494$ mol (approx 0.0049 mol). [1]
(b) Moles HCl = $0.050 \times 0.50 = 0.025$ mol. [1]
(c) Ratio required Mg:HCl is 1:2.
Moles HCl needed for 0.00494 mol Mg = $0.00494 \times 2 = 0.00988$ mol.
Since 0.025 mol HCl is available (> 0.00988), HCl is in excess.
Therefore, Magnesium is the limiting reagent. [2]
(d) Moles H₂ produced = Moles Mg reacted = 0.00494 mol.
Volume H₂ = $0.00494 \times 24.0 = 0.11856$ dm³.
Answer: 0.119 dm³ (or 119 cm³). [2]

12.
(a) Mr Na₂CO₃ = $(2 \times 23.0) + 12.0 + (3 \times 16.0) = 106.0$ g mol⁻¹.
Moles Na₂CO₃ = $1.06 / 106.0 = 0.0100$ mol. [2]
(b) Mass water = $2.86 - 1.06 = 1.80$ g. [1]
(c) Mr H₂O = 18.0 g mol⁻¹.
Moles H₂O = $1.80 / 18.0 = 0.100$ mol. [1]
(d) Ratio H₂O : Na₂CO₃ = $0.100 : 0.0100 = 10 : 1$ .
Therefore, x = 10. [2]

13.
(a) Mr TiCl₄ = $47.9 + (4 \times 35.5) = 189.9$ g mol⁻¹.
Moles TiCl₄ = $10.0 / 189.9 = 0.05266$ mol.
Ratio TiCl₄ : TiO₂ is 1:1.
Moles TiO₂ = 0.05266 mol.
Mr TiO₂ = $47.9 + (2 \times 16.0) = 79.9$ g mol⁻¹.
Mass TiO₂ = $0.05266 \times 79.9 = 4.207$ g.
Answer: 4.21 g. [3]
(b) % Yield = $(\text{Actual} / \text{Theoretical}) \times 100$ .
% Yield = $(3.50 / 4.207) \times 100 = 83.19$ %.
Answer: 83.2%. [2]

14.
(a) Moles H₂SO₄ = $0.025 \times 0.200 = 0.0050$ mol. [1]
(b) Ratio H₂SO₄ : KOH is 1:2.
Moles KOH = $0.0050 \times 2 = 0.010$ mol. [1]
(c) Volume KOH = $20.0 \text{ cm}^3 = 0.020 \text{ dm}^3$ .
Conc KOH = $0.010 / 0.020 = 0.50$ mol dm⁻³. [2]

15.
(a) Mass Cl = $1.90 - 0.48 = 1.42$ g. [1]
(b) Moles Cl atoms = $1.42 / 35.5 = 0.040$ mol. [1]
(c) Formula is MCl₂, so ratio M : Cl is 1 : 2.
Moles M = $0.040 / 2 = 0.020$ mol.
Ar of M = $\text{Mass} / \text{Moles} = 0.48 / 0.020 = 24.0$ .
Answer: 24.0 (Magnesium). [2]

16.
(a) Total moles = $8.00 / 24.0 = 0.3333$ mol. [1]
(b) Eq 1 (Moles): $n_{N2} + n_{O2} = 0.3333$
Eq 2 (Mass): $28n_{N2} + 32n_{O2} = 10.0$ [2]
(c) From Eq 1: $n_{O2} = 0.3333 - n_{N2}$ .
Substitute into Eq 2:
$28n_{N2} + 32(0.3333 - n_{N2}) = 10.0$
$28n_{N2} + 10.666 - 32n_{N2} = 10.0$
$-4n_{N2} = -0.666$
$n_{N2} = 0.1665$ mol.
Mole fraction $X_{N2} = n_{N2} / n_{total} = 0.1665 / 0.3333 = 0.50$ .
Answer: 0.50. [3]

17.
(a) Mass C = $(12.0 / 44.0) \times 0.88 = 0.24$ g. [1]
(b) Mass H = $(2.0 / 18.0) \times 0.54 = 0.06$ g. [1]
(c) Mass O = $0.46 - (0.24 + 0.06) = 0.46 - 0.30 = 0.16$ g. [1]
(d) Moles C = $0.24 / 12.0 = 0.02$ .
Moles H = $0.06 / 1.0 = 0.06$ .
Moles O = $0.16 / 16.0 = 0.01$ .
Ratio C:H:O = $0.02 : 0.06 : 0.01$ .
Divide by smallest (0.01): $2 : 6 : 1$ .
Empirical Formula: C₂H₆O. [3]

18.
(a) Mr Fe₂O₃ = $(2 \times 55.8) + (3 \times 16.0) = 159.6$ g mol⁻¹.
Moles Fe₂O₃ = $10.0 / 159.6 = 0.06266$ mol.
Ratio Al : Fe₂O₃ is 2:1.
Moles Al needed = $0.06266 \times 2 = 0.1253$ mol.
Mass Al = $0.1253 \times 27.0 = 3.38$ g. [3]
(b) Available Al is 5.0 g. Required is 3.38 g.
Al is in excess. Fe₂O₃ is the limiting reagent. [1]

19.
(a) Mr Na₂S₂O₃·5H₂O = $(2 \times 23.0) + (2 \times 32.1) + (3 \times 16.0) + 5(18.0)$
$= 46.0 + 64.2 + 48.0 + 90.0 = 248.2$ g mol⁻¹. [1]
(b) Moles salt = $12.4 / 248.2 = 0.04996$ mol ≈ 0.0500 mol.
Volume = $250 \text{ cm}^3 = 0.250 \text{ dm}^3$ .
Concentration = $0.0500 / 0.250 = 0.200$ mol dm⁻³. [2]

20.
(a) Moles of gas X = $1.00 / 24.0 = 0.04167$ mol.
Molar Mass = $\text{Mass} / \text{Moles} = 1.75 / 0.04167 = 42.0$ g mol⁻¹. [2]
(b) Empirical formula mass of CH₂ = $12.0 + 2.0 = 14.0$ .
Ratio = $42.0 / 14.0 = 3$ .
Molecular Formula = C₃H₆. [1]