1.3.3 Reacting Volumes

In chemical reactions involving gases or solutions, we often need to calculate the volumes of reactants and products. This is particularly important in reactions where gases are involved, or when dealing with solutions of known concentration.

Calculating Volumes of Gases

At room temperature and pressure (RTP), 1 mole of any gas occupies 24 dm³. This relationship can be used to calculate the volume of gases involved in a reaction.

Formula:

\text{Volume of gas (dm³)} = \text{Moles of gas} \times 24 \, \text{dm}^3

This is true for all gases at RTP, regardless of their identity, as long as the temperature and pressure remain constant.

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Example: Volume of Oxygen Required to React with Hydrogen Question: What volume of oxygen gas (O₂) is required to react with hydrogen gas (H₂) to form 48 dm³ of water vapour?

Step 1: Write the balanced chemical equation

2H_2 + O_2 \rightarrow 2H_2O

Step 2: Use the mole ratio From the balanced equation, 2 moles of hydrogen produce 2 moles of water. Therefore, 48 dm³ of water vapor corresponds to 48 dm³ of hydrogen gas.

The mole ratio between hydrogen and oxygen is 2:1, so for every 2 moles (or 48 dm³) of hydrogen, 1 mole (or 24 dm³) of oxygen is required.

Thus, the volume of O₂ required is 24 dm³.

Calculating Volumes of Solutions

When dealing with solutions, we can calculate the volume required for a reaction if we know the concentration and the number of moles of the solute involved.

Formula:

\text{Moles of solute} = \text{Concentration} \times \text{Volume (dm³)}

Where:

Concentration is in mol/dm³ (sometimes denoted as M for molarity).
Volume is in dm³ (1 dm³ = 1000 cm³). To calculate the volume:

\text{Volume (dm³)} = \frac{\text{Moles}}{\text{Concentration (mol/dm³)}}

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Example: Volume of Sodium Hydroxide Solution Required for Neutralisation Question: How much 0.2 mol/dm³ sodium hydroxide (NaOH) solution is needed to neutralise 50 cm³ of 0.1 mol/dm³ hydrochloric acid (HCl)?

Step 1: Write the balanced chemical equation

NaOH + HCl \rightarrow NaCl + H_2O

Step 2: Calculate the moles of HCl Using the concentration-volume relationship:

\text{Moles of HCl} = \text{Concentration} \times \frac{\text{Volume}}{1000}

= 0.1 \, \text{mol/dm}^3 \times \frac{50 \, \text{cm}^3}{1000}

= :highlight[0.005 \, \text{mol}]

Step 3: Use the mole ratio The mole ratio between NaOH and HCl is 1:1, so the moles of NaOH required = 0.005 mol.

Step 4: Calculate the volume of NaOH: Using the concentration of the NaOH solution:

\text{Volume of NaOH (dm³)} = \frac{\text{Moles of NaOH}}{\text{Concentration}}

= \frac{0.005 \, \text{mol}}{0.2 \, \text{mol/dm}^3} = 0.025 \, \text{dm}^3

Convert to cm³:

0.025 \, \text{dm}^3 = :highlight[25 \, \text{cm}^3]

Thus, 25 cm³ of sodium hydroxide solution is needed.

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Summary

For gases, use the relationship that 1 mole of gas = 24 dm³ at RTP to calculate gas volumes.
For solutions, use the formula ( $\text{Moles} = \text{Concentration} \times \text{Volume (dm³)}$ ) to find the volume required for reactions in solution.
Balanced equations are key for identifying the correct mole ratios and applying them to volume calculations for both gases and solutions.

Reacting Volumes Simplified Revision Notes for A-Level AQA Chemistry

1.3.3 Reacting Volumes

Calculating Volumes of Gases

Calculating Volumes of Solutions

Summary

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