ASChemistryAQARevision NotesPhysical ChemistryThe Mole, Avogadro & The Ideal Gas EquationReacting Volumes

Reacting Volumes (AQA AS Chemistry): Revision Note

Stewart Hird

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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) = Alternative text not available

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

  • A 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 cm3 to dm3 

  • 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) = fraction numerator 2.5 space straight g over denominator 100 space straight g space mol to the power of negative 1 end exponent end fraction = 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) = fraction numerator amount space open parentheses mol close parentheses over denominator concentration space open parentheses mol space dm to the power of negative 3 end exponent close parentheses end fraction

    • Volume (HCl) = fraction numerator 0.05 space mol over denominator 1.0 space mol space dm to the power of negative 3 end exponent end fraction = 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] = fraction numerator amount space open parentheses mol close parentheses over denominator volume space open parentheses dm cubed close parentheses end fraction

    • [HCl] = begin mathsize 14px style fraction numerator 0.00250 over denominator 0.0200 end fraction end style = 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) = fraction numerator bold volume bold space bold of bold space bold gas bold space stretchy left parenthesis dm cubed stretchy right parenthesis over denominator bold 24 bold. bold 0 end fraction

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.

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