Pearson Edexcel International GCSE (9–1) · CHEMISTRY · Chapter
Chemical Formulae, Equations and Calculations Part 2
144 practice questions
Topics covered
- Calculations involving gas volumes19 practice questions
- Calculating the volume of a certain number of moles19 practice questions
- Using the molar volume in calculations with chemical equations19 practice questions
- Concentrations of solutions19 practice questions
- Calculations with equations involving solutions19 practice questions
- The standard calculation19 practice questions
- Reversing the calculations20 practice questions
Practice questions
51–60 of 144
Question 51
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 52
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 53
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 54
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.
Question 55
2 marks · Hard
Answer and see the solution →A student collects 8.4 g of oxygen gas at room temperature and pressure (rtp). Calculate the volume, in dm³, of this oxygen gas. (Ar: O = 16; molar volume = 24 dm³ at rtp.)
Question 56
5 marks · Hard
Answer and see the solution →When sodium nitrate (ce{NaNO3}) decomposes on heating, it produces sodium nitrite (ce{NaNO2}) and oxygen gas (ce{O2}). The equation is:
(a)[2 marks]Calculate the number of moles in 0.34 g of sodium nitrate. (Ar: Na = 23, N = 14, O = 16)
(b)[3 marks]Use your answer to (a) to calculate the volume, in cm³, of oxygen gas produced at rtp.
(Molar volume = 24,000 cm³ at rtp.)
Question 57
2 marks · Hard
Answer and see the solution →Calculate the number of moles of neon gas present in a 4.8 dm³ flask at room temperature and pressure. Give your answer to two decimal places. (Molar volume = 24 dm³ at rtp.)
Question 58
1 mark · Easy
Answer and see the solution →State the units used for concentration when it is measured in moles per cubic decimetre.
Question 59
3 marks · Easy
Answer and see the solution →(a)[1 mark]Give the formula that links concentration (in mol/dm³), amount of substance (in mol), and volume of solution (in dm³).
(b)A student dissolves 0.080 mol of potassium nitrate (KNO₃) in enough water to make 0.40 dm³ of solution.
(i)[2 marks]Calculate the concentration of this potassium nitrate solution in mol/dm³.
Question 60
2 marks · Easy
Answer and see the solution →Calculate the amount of substance (in mol) in 200 cm³ of a solution with a concentration of 0.150 mol/dm³. Give your answer to 2 decimal places.