Practise Volumes of gases. 15 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy4 marks
The volume of one mole of any gas at room temperature and pressure (rtp) is 24 dm3. volume of gas at rtp in dm3 = amount of gas in mol × 24
(a) Calculate the volume of 0.50 mol of oxygen at rtp.[1]
(b) Calculate the amount, in moles, of carbon dioxide in 6.0 dm3 of the gas at rtp.[2]
(c) Which sample of gas has the largest volume at rtp? Tick (✓) one box.[1]
1 mol of hydrogen
1 mol of oxygen
1 mol of carbon dioxide
They all have the same volume
Show the answer and mark scheme
(a)Answer: 12 dm3
12 (dm3)
(b)Answer: 0.25 mol
6.0 ÷ 24
0.25 (mol)
(c)Answer: They all have the same volume
Question 2Medium5 marks
The volume of one mole of any gas at room temperature and pressure (rtp) is 24 dm3. Relative atomic masses (Ar): H = 1, C = 12, N = 14, O = 16
(a) Calculate the volume of 8.8 g of carbon dioxide at rtp.[2]
(b) Calculate the mass of 600 cm3 of ammonia, NH3, at rtp.[3]
Show the answer and mark scheme
(a)Answer: 4.8 dm3
moles of CO2 = 8.8 ÷ 44 = 0.2
0.2 × 24 = 4.8 (dm3)
(b)Answer: 0.425 g
600 cm3 = 0.600 dm3 / 24 dm3 = 24 000 cm3
moles of NH3 = 0.600 ÷ 24 = 0.025
0.025 × 17 = 0.425 (g)
Question 3Hard7 marks
A student reacted a piece of zinc with excess dilute sulfuric acid and collected the hydrogen produced in a gas syringe. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) The volume of one mole of any gas at rtp is 24 dm3. Relative atomic mass (Ar): Zn = 65
(a) The student collected 96.0 cm3 of hydrogen at rtp. Calculate the mass of zinc that reacted.[3]
(b) The piece of zinc had a mass of 0.30 g. Calculate the volume of hydrogen that 0.30 g of zinc should produce at rtp. Give your answer in cm3 to 3 significant figures.[2]
(c) Suggest two reasons why the student collected less hydrogen than expected.[2]
Show the answer and mark scheme
(a)Answer: 0.26 g
moles of H2 = 96.0 ÷ 24 000 = 0.00400
moles of Zn = 0.00400
0.00400 × 65 = 0.26 (g)
(b)Answer: 111 cm3
moles of Zn = 0.30 ÷ 65 = 0.00462
0.00462 × 24 000 = 111 (cm3)
(c)
some hydrogen escaped before the bung / syringe was connected
the apparatus leaked
the zinc was impure / had an oxide coating
the reading was taken before all of the zinc had reacted