Moles and Gases

This section explains moles and gases, covering volumes of gases, gas volumes and equations and using moles to balance equations. 

Volume of Gases

In chemistry, gases occupy a fixed volume at a given temperature and pressure. Under room temperature and pressure (RTP) (20°C and 1 atmosphere), one mole of any gas occupies 24 dm³.

Formula for Gas Volume

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

or rearranged:

$$\text{Moles of gas} = \frac{\text{Volume of gas (dm}^3)}{24}$$

Example Calculation

Calculate the volume occupied by 2.5 moles of oxygen (O₂) at RTP.

$$\text{Volume} = 2.5 \times 24 = 60 \text{ dm}^3$$

If 48 dm³ of carbon dioxide (CO₂) is produced in a reaction, the number of moles is:

$$\text{Moles} = \frac{48}{24} = 2 \text{ mol}$$ 

Gas Volumes and Equations

In reactions involving gases, we can use the mole ratio from balanced equations to find the volume of gases involved.

Example: Combustion of Methane

$$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$$ 

From the equation:

• 1 mole of methane reacts with 2 moles of oxygen.
• This means 1 volume of methane reacts with 2 volumes of oxygen.
• It produces 1 volume of carbon dioxide and 2 volumes of steam.

If 12 dm³ of methane is burned:

• Oxygen required: $12 \times 2 = 24\text{dm}^3$
• Carbon dioxide produced: 12 dm³

Since water is a gas at RTP, water vapour = 24 dm³.

Using Moles to Balance Equations

Moles can help balance chemical equations by ensuring the number of atoms on both sides is equal.

Example: Formation of Ammonia

$$\text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3$$ ​ 

Step 1: Count atoms.

• 1 N₂ molecule has 2 nitrogen atoms.
• 1 NH₃ molecule has 1 nitrogen atom, so we need 2 NH₃ molecules.

$$\text{N}_2 + \text{H}_2 \rightarrow 2\text{NH}_3$$

Step 2: Balance hydrogen.

• 2 NH₃ molecules contain 6 hydrogen atoms.
• Each H₂ molecule has 2 hydrogen atoms, so we need 3 H₂ molecules.

Final balanced equation:

$$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$$ 

Example: Combustion of Propane

$$\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$$ 

Step 1: Balance carbon.

• 1 C₃H₈ contains 3 carbon atoms, so we need 3 CO₂ molecules.

$$\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + \text{H}_2\text{O}$$ 

Step 2: Balance hydrogen.

• 1 C₃H₈ has 8 hydrogen atoms, so we need 4 H₂O molecules.

$$\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$$ 

Step 3: Balance oxygen.

• 3 CO₂ molecules = 6 oxygen atoms.
• 4 H₂O molecules = 4 oxygen atoms.
• Total oxygen atoms required = 10, so we need 5 O₂ molecules.

Final balanced equation:

$$\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$$ 

Key Points to Remember

• At RTP, 1 mole of any gas = 24 dm³.
• Volume (dm³) = moles × 24.
• Use mole ratios from balanced equations to calculate gas volumes.
• Balancing equations ensures the number of atoms is equal on both sides.

This guide provides a solid foundation for quantitative chemistry involving gases and moles.