Pearson Edexcel International GCSE (9–1) · CHEMISTRY · Section
Using the molar volume in calculations with chemical equations
19 practice questions
Practice questions
1–8 of 8 · Normal
Question 1
2 marks · Normal
Answer and see the solution →Calculate the volume, in cm³, of 0.16 g of oxygen gas (O₂) at room temperature and pressure. (Ar: O = 16; molar volume = 24,000 cm³ at rtp)
Question 2
4 marks · Normal
Answer and see the solution →Hydrochloric acid reacts with sodium carbonate to produce sodium chloride, carbon dioxide and water. The balanced equation is:
(a)[2 marks]Calculate the number of moles in 1.06 g of sodium carbonate. (Ar: Na = 23, C = 12, O = 16)
(b)[2 marks]Calculate the volume, in dm³, of carbon dioxide produced at rtp by this mass of sodium carbonate. (Molar volume at rtp = 24 dm³)
Question 3
3 marks · Normal
Answer and see the solution →Calculate the number of moles in 360 cm³ of nitrogen gas at room temperature and pressure. Give your answer to two decimal places. (Molar volume = 24,000 cm³ at rtp)
Question 4
3 marks · Normal
Answer and see the solution →A student needs to calculate the mass of carbon dioxide produced when 48,000 cm³ of carbon dioxide is collected at room temperature and pressure. Explain the steps, including the formulae needed, to calculate the mass of carbon dioxide produced. (Ar: C = 12, O = 16; molar volume = 24,000 cm³ at rtp)
Question 5
4 marks · Normal
Answer and see the solution →The table below shows the volumes, in cm³, of a gas collected from several reactions at room temperature and pressure. Fill in the number of moles of gas for each row, using the molar volume of 24,000 cm³ at rtp.
Volume (cm³) Moles 6000 12,000 1800 24,000 Question 6
1 mark · Normal
Answer and check →Which calculation gives the number of moles of a gas from its volume at rtp (using cm³)?
- A.moles = volume / 24,000
- B.moles = mass / Mr
- C.moles = volume × 24,000
- D.moles = Mr / mass
Question 7
2 marks · Normal
Answer and see the solution →Calculate the volume, in dm³, of chlorine gas produced at rtp from the complete decomposition of 2.7 g of potassium chlorate, KClO₃. The balanced equation is:
(Ar: K = 39, Cl = 35.5, O = 16; molar volume = 24 dm³)
Question 8
4 marks · Normal
Answer and see the solution →Ammonium nitrate decomposes to form dinitrogen oxide (N2O) and water. The balanced equation is:
A practical uses 8.0 g of ammonium nitrate. (Ar: N = 14, H = 1, O = 16; molar volume at rtp = 24 dm³)
(a)[2 marks]Calculate the number of moles of ammonium nitrate used.
(b)[2 marks]Calculate the volume, in dm³, of dinitrogen oxide (N₂O) formed at rtp.