Home / iGCSE Chemistry 0620 Extended – 3.2 Relative masses of atoms and molecules- Exam Style Questions Paper 2

iGCSE Chemistry 0620 Extended - 3.2 Relative masses of atoms and molecules- Exam Style Questions Paper 2- New Syllabus

iGCSE Chemistry MCQ (Extended) Paper 2 - All Chapters

Question

The equation for the reaction between copper(II) oxide, CuO, and ammonia is given.

\[3\mathrm{CuO} + 2\mathrm{NH_3} \rightarrow 3\mathrm{Cu} + 3\mathrm{H_2O} + \mathrm{N_2}\]

[M: CuO, 80; NH₃, 17; H₂O, 18; N₂, 28]
[A: Cu, 64]

Which statements about this reaction are correct?

  1. 80.0g of copper(II) oxide reacts exactly with 17.0g of ammonia.
  2. 4.0g of copper(II) oxide reacts with excess ammonia to produce 3.2g of copper.
  3. The mass of water produced in this reaction is greater than the mass of nitrogen.

A. 1, 2 and 3 
B. 1 and 2 only 
C. 1 and 3 only 
D. 2 and 3 only

▶️ Answer/Explanation

Correct Answer: D (2 and 3 only)

Calculations:
Statement 1: 3 mol CuO (3×80=240g) reacts with 2 mol NH₃ (2×17=34g). So 80g CuO reacts with (80/240)×34 = 11.33g NH₃, not 17g. ❌
Statement 2: 3 mol CuO (240g) produces 3 mol Cu (3×64=192g). So 4g CuO produces (4/240)×192 = 3.2g Cu. ✅
Statement 3: 3 mol H₂O (3×18=54g) and 1 mol N₂ (28g). 54g > 28g, so mass of H₂O > mass of N₂. ✅

✅ Statements 2 and 3 are correct.

Question

The equation for the reaction of magnesium with dilute sulfuric acid is shown.

$$\text{Mg} + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + \text{H}_2$$

[$M_r$: $\text{MgSO}_4 = 120$]

Which mass of magnesium sulfate is formed when 12 g of magnesium completely reacts with dilute sulfuric acid?

A. 5 g
B. 10 g
C. 60 g
D. 120 g

▶️ Answer/Explanation
To determine the mass of magnesium sulfate produced, we first calculate the number of moles of magnesium used in the reaction. Given the molar mass of magnesium is 24 g/mol, a 12 g sample corresponds to exactly 0.5 moles. According to the balanced chemical equation, magnesium reacts with sulfuric acid to form magnesium sulfate in a 1:1 molar ratio, meaning 0.5 moles of magnesium will yield 0.5 moles of $\text{MgSO}_4$. The relative formula mass ($M_r$) of $\text{MgSO}_4$ is given as 120. Multiplying the moles of product by its molar mass ($0.5 \times 120$) gives a final theoretical yield of 60 g. This precisely matches the expected stoichiometric proportions.
✅ Answer: (C)

Question

Which sample contains the largest number of molecules?

A. 16 g of methane, $\text{CH}_4(\text{g})$
B. 16 g of oxygen, $\text{O}_2(\text{g})$
C. 16 g of phosphorus, $\text{P}_4(\text{s})$
D. $16\text{ dm}^3$ of methane at r.t.p., $\text{CH}_4(\text{g})$

▶️ Answer/Explanation
Let’s compute the moles of molecules in each sample, since moles are directly proportional to the total count of molecules:
– A: Moles of $\text{CH}_4 = \frac{16\text{ g}}{16\text{ g/mol}} = 1.0\text{ mol}$
– B: Moles of $\text{O}_2 = \frac{16\text{ g}}{32\text{ g/mol}} = 0.5\text{ mol}$
– C: Moles of $\text{P}_4 = \frac{16\text{ g}}{4 \times 31\text{ g/mol}} = 0.13\text{ mol}$
– D: Moles of $\text{CH}_4 = \frac{16\text{ dm}^3}{24\text{ dm}^3\text{/mol}} = 0.67\text{ mol}$
Sample A contains exactly $1.0\text{ mole}$ of molecules, which is clearly the highest.
✅ Answer: (A)
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