Secondary 3 Chemistry Stoichiometry Moles Quiz

Secondary 3 Chemistry Quiz - Stoichiometry Moles

Answer Key

Section A: Multiple Choice & Short Answer

1. (a) mol⁻¹
[1 mark]
Common mistake: Students confuse the unit of Avogadro constant (mol⁻¹) with the unit of amount of substance (mol).

2. (d) 4 mol
[1 mark]
Each molecule of H₂SO₄ contains 4 oxygen atoms, so 1 mol of H₂SO₄ contains 4 mol of O atoms.

3. (c) 100 g mol⁻¹
[1 mark]
Mr = 40 + 12 + (3 × 16) = 100

4. (a) 1 g of H₂
[1 mark]
H₂ has the smallest molar mass (2 g mol⁻¹), so 1 g contains the most moles and therefore the most molecules.

5. (a) 12 dm³
[1 mark]
At rtp, molar volume = 24 dm³ mol⁻¹. Volume = 0.5 × 24 = 12 dm³.

6. A mole is the amount of substance that contains as many particles (atoms, molecules, or ions) as there are atoms in exactly 12 g of carbon-12.
[1 mark]
Award 1 mark for a clear, correct definition. Accept "Avogadro number of particles" as part of the definition.

7. 6.02 × 10²³ mol⁻¹
[1 mark]
Accept 6.02 × 10²³ or 6.022 × 10²³.

8. Molar mass is the mass of one mole of a substance. It is numerically equal to the relative atomic mass (or relative molecular mass) expressed in grams per mole (g mol⁻¹).
[1 mark]
Award 1 mark for a correct definition including the unit g mol⁻¹ or equivalent.

9. 24 dm³ mol⁻¹
[1 mark]
Accept 24 dm³ or 24 000 cm³.

10. Molar mass of H₂O = (2 × 1) + 16 = 18 g mol⁻¹
Number of moles = 36 / 18 = 2.0 mol
[1 mark]
Award 1 mark for the correct answer. No working required for 1 mark.

Section B: Calculations

11.

(a) Mr of NaOH = 23 + 16 + 1 = 40
Number of moles = 4.0 / 40 = 0.10 mol
[2 marks]

• 1 mark for correct Mr
• 1 mark for correct answer
  Common mistake: Using wrong Mr value.

(b) Mr of NH₄NO₃ = 14 + (4 × 1) + 14 + (3 × 16) = 80
Number of moles = 132 / 80 = 1.65 mol
[2 marks]

• 1 mark for correct Mr
• 1 mark for correct answer
  Common mistake: Forgetting to count all atoms in the formula (e.g., calculating N₂H₄O₃ incorrectly).

12.

(a) Mr of MgO = 24 + 16 = 40
Mass = 0.25 × 40 = 10 g
[2 marks]

• 1 mark for correct Mr
• 1 mark for correct answer

(b) Mr of CO₂ = 12 + (2 × 16) = 44
Mass = 3.0 × 44 = 132 g
[2 marks]

• 1 mark for correct Mr
• 1 mark for correct answer

13. Mr of CO₂ = 12 + (2 × 16) = 44
Number of moles of CO₂ = 4.4 / 44 = 0.10 mol
Number of molecules = 0.10 × 6.02 × 10²³ = 6.02 × 10²² molecules
[3 marks]

• 1 mark for correct Mr
• 1 mark for correct moles
• 1 mark for correct number of molecules
  Common mistake: Forgetting to multiply by Avogadro constant, or using the wrong power of 10.

14.

(a) Number of moles = 36 / 24 = 1.5 mol
[2 marks]

• 1 mark for using molar volume (24 dm³ mol⁻¹)
• 1 mark for correct answer

(b) Mr of O₂ = 2 × 16 = 32
Mass = 1.5 × 32 = 48 g
[2 marks]

• 1 mark for correct Mr
• 1 mark for correct answer
  Award 1 mark for correct method even if part (a) was wrong (error carried forward).

15.

Empirical formula = CH₂O
[4 marks]

• 1 mark for dividing each % by correct Ar
• 1 mark for calculating ratios
• 1 mark for simplifying to whole numbers
• 1 mark for correct empirical formula
  Common mistake: Not dividing by the smallest ratio, or rounding incorrectly (e.g., 3.33 → 3 instead of keeping as 1 when divided).

16.

(a) Number of moles of Mg = 2.4 / 24 = 0.10 mol
[2 marks]

• 1 mark for using correct Ar
• 1 mark for correct answer

(b) From the equation: Mg : H₂ = 1 : 1
Number of moles of H₂ = 0.10 mol
[2 marks]

• 1 mark for correct mole ratio
• 1 mark for correct answer

(c) Volume of H₂ = 0.10 × 24 = 2.4 dm³
[2 marks]

• 1 mark for using molar volume
• 1 mark for correct answer
  Award error carried forward from (b).

Section C: Structured & Applied Questions

17.

(a) Number of moles = concentration × volume (in dm³)
= 0.2 × (500/1000) = 0.2 × 0.5 = 0.10 mol
[2 marks]

• 1 mark for correct formula
• 1 mark for correct answer
  Common mistake: Not converting cm³ to dm³ (using 500 instead of 0.5).

(b) Mr of NaCl = 23 + 35.5 = 58.5
Mass = 0.10 × 58.5 = 5.85 g
[2 marks]

• 1 mark for correct Mr
• 1 mark for correct answer

(c) Steps:

1. Weigh out 5.85 g of sodium chloride using an electronic balance.
2. Transfer the sodium chloride into a beaker and add some distilled water. Stir to dissolve.
3. Transfer the solution into a 500 cm³ volumetric flask. Rinse the beaker and stirring rod with distilled water and add the washings to the flask.
4. Add distilled water until the bottom of the meniscus reaches the 500 cm³ mark on the flask.
5. Stopper the flask and invert several times to mix thoroughly.
   [2 marks]

• 1 mark for weighing correct mass and dissolving
• 1 mark for using volumetric flask and making up to the mark
  Award partial credit for partially correct descriptions.

18.

(a) Mr of CaCO₃ = 40 + 12 + (3 × 16) = 100
Number of moles = 10 / 100 = 0.10 mol
[2 marks]

• 1 mark for correct Mr
• 1 mark for correct answer

(b) From the equation: CaCO₃ : CO₂ = 1 : 1
Number of moles of CO₂ = 0.10 mol
Volume of CO₂ = 0.10 × 24 = 2.4 dm³
[3 marks]

• 1 mark for correct mole ratio
• 1 mark for correct moles of CO₂
• 1 mark for correct volume

(c) Percentage yield = (actual yield / theoretical yield) × 100%
= (4.0 / 2.4) × 100% = 166.7%
[2 marks]

• 1 mark for correct formula
• 1 mark for correct calculation
  Note: A yield > 100% suggests impurities or incomplete drying. Accept the calculated value. Common mistake: Inverting the fraction (2.4/4.0 instead of 4.0/2.4).

19.

(a)

Empirical formula = CO₂
[3 marks]

• 1 mark for dividing by Ar
• 1 mark for calculating ratios
• 1 mark for correct empirical formula

(b) Mr of empirical formula (CO₂) = 12 + (2 × 16) = 44
n = molar mass / empirical formula mass = 88 / 44 = 2
Molecular formula = (CO₂)₂ = C₂O₄
[3 marks]

• 1 mark for calculating empirical formula mass
• 1 mark for finding n = 2
• 1 mark for correct molecular formula
  Note: C₂O₄ is the correct answer based on the data given. In practice, this could represent a dimer or a different compound. Accept the calculated answer.

20.

(a) From the equation: N₂ : H₂ = 1 : 3
At rtp, volume ratio = mole ratio = 1 : 3
Volume of N₂ needed = 18 / 3 = 6 dm³
[3 marks]

• 1 mark for correct mole/volume ratio from equation
• 1 mark for correct substitution
• 1 mark for correct answer
  Common mistake: Using the ratio the wrong way (multiplying instead of dividing).

(b) From the equation: H₂ : NH₃ = 3 : 2
Number of moles of H₂ = 18 / 24 = 0.75 mol
Number of moles of NH₃ = (2/3) × 0.75 = 0.50 mol
Mr of NH₃ = 14 + (3 × 1) = 17
Mass of NH₃ = 0.50 × 17 = 8.5 g
[4 marks]

• 1 mark for moles of H₂
• 1 mark for correct mole ratio and moles of NH₃
• 1 mark for correct Mr of NH₃
• 1 mark for correct final answer
  Award error carried forward throughout.

End of Answer Key