Reacting Volumes (AQA A Level Chemistry): Revision Note

Exam code: 7405

Stewart Hird

Written by: Stewart Hird

Reviewed by: Caroline Carroll

Updated on

Volumes & Concentrations of Solutions

  • The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dm3 of  solution

    • The solute is the substance that dissolves in a solvent to form a solution

    • The solvent is often water

concentration (mol dm-3) = number of moles of solute (mol)volume of solution (dm3)

  • concentrated solution is a solution that has a high concentration of solute

  • dilute solution is a solution with a low concentration of solute

  • When carrying out calculations involve concentrations in mol dm-3 the following points need to be considered:

    • Change mass in grams to moles

    • Change cmto dm

  • To calculate the mass of a substance present in solution of known concentration and volume:

    • Rearrange the concentration equation

number of moles (mol) = concentration (mol dm-3) x volume (dm3)

  • Multiply the moles of solute by its molar mass

mass of solute (g) = number of moles (mol) x molar mass (g mol-1)

Worked Example

Calculating volume from concentration

Calculate the volume of 1.0 mol dm-3 hydrochloric acid required to completely react with 2.5 g of calcium carbonate.

Answer:

  1. Write the balanced symbol equation

    • CaCO3  +  2HCl  →  CaCl2  +  H2O  +  CO2

  2. Calculate the amount, in moles, of calcium carbonate:

    • n(CaCO3) = 2.5 g100 g mol1 = 0.025 mol

  3. Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry:

    • 1 mol of CaCO3 requires 2 mol of HCl

    • So 0.025 mol of CaCO3 requires 0.05 mol of HCl

  4. Calculate the volume of HCl required:

    • Volume (HCl) = amount (mol)concentration (mol dm3)

    • Volume (HCl) = 0.05 mol1.0 mol dm3 = 0.05 dm3

    • So, the volume of hydrochloric acid required is 0.05 dm3 

Worked Example

Neutralisation calculation

25.0 cm3 of 0.050 mol dm-3 sodium carbonate solution was completely neutralised by 20.0 cm3 of dilute hydrochloric acid.

Calculate the concentration, in mol dm-3, of the hydrochloric acid.

Answer:

  1. Write the balanced symbol equation:

    • Na2CO3  +  2HCl  →  Na2Cl2  +  H2O  +  CO2

  2. Calculate the amount, in moles, of sodium carbonate reacted

    • n(Na2CO3) = 0.025 dm3 x 0.050 mol dm-3

    • n(Na2CO3) = 0.00125 mol

  3. Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry:

    • 1 mol of Na2CO3 reacts with 2 mol of HCl, so the molar ratio is 1 : 2

    • Therefore, 0.00125 moles of Na2CO3 react with 0.00250 moles of HCl

  4. Calculate the concentration, in mol dm-3 of hydrochloric acid:

    • [HCl] = amount (mol)volume (dm3)

    • [HCl] = 0.002500.0200 = 0.125 mol dm-3

Volumes of gases

  • Avogadro suggested that ‘equal volumes of gases contain the same number of molecules’ (also called Avogadro’s hypothesis)

  • At room temperature and pressure, one mole of any gas has a volume of 24.0 dm3 

    • Room temperature is 20 oC

    • Room pressure is 1 atmosphere 

  • Using the following equations, the molar gas volume, 24.0 dm3, can be used to find:

    • The volume of a given mass or number of moles of gas

    • The mass or number of moles of a given volume of gas

volume of gas (dm3) = amount of gas (mol) x 24.0

amount of gas (mol) = volume of gas (dm3)24.0

Worked Example

Calculating the volume of gas

Complete the table to calculate the volume that the gases occupy:

Gas

Amount of gas (mol)

Volume of gas (dm3)

Hydrogen

3.0

 

Carbon dioxide

0.25

 

Oxygen 

5.4

 

Ammonia

0.02

 

Answers:

Gas

Amount of gas (mol)

Volume of gas (dm3)

Hydrogen

3.0

3.0 x 24.0 = 72.0

Carbon dioxide

0.25

0.25 x 24.0 = 6.0

Oxygen 

5.4

5.4 x 24.0 = 129.6

Ammonia

0.02

0.02 x 24.0 = 0.48

Worked Example

Calculating the number of moles of gas

Complete the table to calculate the number of moles of gas:

Gas

Amount of gas (mol)

Volume of gas (dm3)

Methane

 

225.6

Carbon monoxide

 

7.2

Sulfur dioxide 

 

960

Answers:

Gas

Amount of gas (mol)

Volume of gas (dm3)

Methane

225.6 / 24.0 = 9.43.0

225.6

Carbon monoxide

7.2 / 24.0 = 0.30

7.2

Sulfur dioxide 

960 / 24.0 = 40

960

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Stewart Hird

Author: Stewart Hird

Expertise: Chemistry Content Creator

Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.