Question 1Multiple choice · 1 markA* challenge
CaCO3 → CaO + CO2
What is the maximum mass of CaO that can be made from 16.8 g of CaCO3?
- A 4.70 g
- B 9.41 g
- C 18.8 g
- D 30.0 g
Show answer
Answer: B
Free Relative masses & the mole questions for IGCSE Chemistry (0620), part of Stoichiometry. Each question has the answer and the marking points an examiner would look for.
Generate a full Relative Masses & the Mole worksheet →All Stoichiometry questionsCaCO3 → CaO + CO2
What is the maximum mass of CaO that can be made from 16.8 g of CaCO3?
Answer: B
A compound contains 81.8% C, 18.2% H by mass.
What is its empirical formula?
Answer: C
Calcium carbonate decomposes on heating: CaCO₃ → CaO + CO₂. (Mr CaCO₃ = 100, CaO = 56)
Calculate the mass of calcium oxide made from 25 g of calcium carbonate.
Calculate the volume of carbon dioxide produced at r.t.p.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 14 g moles = 25 ÷ 100 C1 × 56 = 14 g A1 | 2 |
| (b) | 6 dm³ moles × 24 C1 6 dm³ A1 | 2 |
This question is about calculating formulae and amounts of substance.
State what is meant by one mole of a substance.
A compound Q contains 30.4% nitrogen and 69.6% oxygen by mass.
Calculate the empirical formula of Q. [Aᵣ: N, 14; O, 16]
The relative molecular mass of Q is 92. Use your answer to (b)(i) to deduce the molecular formula of Q.
Calculate the volume of hydrogen, in cm³ at r.t.p., produced when 0.24 g of magnesium reacts with excess hydrochloric acid. [Aᵣ: Mg, 24]
8.00 g of impure calcium carbonate is heated strongly until it decomposes completely. 1.73 dm³ of carbon dioxide is collected at r.t.p. Calculate the percentage purity of the calcium carbonate. [Mᵣ: CaCO₃, 100]
Explain why equal volumes of different gases at the same temperature and pressure contain the same number of molecules, but have different masses.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 6.02 × 10²³ particles the amount containing 6.02 × 10²³ particles (Avogadro constant) B1 | 1 |
| (b)(i) | NO₂ 30.4/14 = 2.17 and 69.6/16 = 4.35 M1 ratio 1.00 : 2.00 M1 NO₂ A1 | 3 |
| (ii) | N₂O₄ Mᵣ of empirical formula and ratio 92/its Mᵣ M1 N₂O₄ A1 | 2 |
| (c) | 240 cm³ moles Mg = 0.0100 = moles H₂ M1 240 cm³ A1 | 2 |
| (d) | 90% moles CO₂ = 1.73/24 = 0.0720 M1 mass of CaCO₃ = 7.20 g M1 90% A1 | 3 |
| (e) | same number of moles; different Mr one mole of any gas occupies the same volume (24 dm³ at r.t.p.) B1 each gas has a different Mᵣ so the same number of moles has a different mass B1 | 2 |
Magnesium burns in oxygen: 2Mg + O₂ → 2MgO. A student burns 7.2 g of magnesium. (Ar: Mg = 24, O = 16)
Calculate the number of moles of magnesium.
Calculate the maximum mass of magnesium oxide formed.
The student obtained a smaller mass. Calculate the percentage yield if she obtained 10.2 g.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 0.3 mol 0.3 mol B1 | 1 |
| (b) | 12 g moles MgO = moles Mg C1 × 40 = 12 g A1 | 2 |
| (c) | 85 % actual ÷ theoretical × 100 C1 85 % A1 | 2 |
Calcium carbonate reacts with hydrochloric acid: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. (Mr CaCO₃ = 100; molar gas volume = 24 dm³ at r.t.p.)
Calculate the mass of calcium carbonate needed to produce 720 cm³ of carbon dioxide at r.t.p.
Define the term molar gas volume.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 3 g moles CO₂ = 720 ÷ 24 000 C1 moles CaCO₃ = moles CO₂ C1 3 g A1 | 3 |
| (b) | volume of 1 mol gas at rtp = 24 dm³ the volume occupied by one mole of any gas (24 dm³ at room temperature and pressure) B1 | 1 |
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