Excess Calculations - Reacting Gas Volumes

Chemistry

Excess Calculations - Reacting Gas Volumes

Introduction:

• One mole of any gas occupies a similar volume at the same temperature and pressure.
• Equal volumes of gases contain the same number of moles of gas.
• A balanced chemical equation reflects this principle, as it shows the same number of moles for reactants and products.

Examples from Balanced Equations

N₂(g) + 3H₂(g) → 2NH₃(g)

• 1 mole N₂ reacts with 3 moles H₂ to produce 2 moles NH₃.
• This corresponds to 1 litre of N₂ reacting with 3 litres of H2 to produce 2 litres of NH₃.
• Or 10 cm³ N2 reacts with 30 cm³ H₂ to produce 20 cm³ NH₃.

C(s) + O₂(g) → CO₂(g)

• 1 mole C reacts with 1 mole O₂ to produce 1 mole CO₂.
• This corresponds to 1 litre C reacting with 1 litre O2 to produce 1 litre CO₂.
• Or 10 cm³ C reacts with 10 cm³ O₂ to produce 10 cm³ CO₂.

Exclusions

• Liquid or solid reactants and products are not included in these volume calculations.

Worked Example

• Given: 30 cm³ CH₄ completely burned in 100 cm³ O₂.
• Reaction: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
• Step 1: Determine the excess reactant.
  • 1 mole CH₄ reacts with 2 moles O₂.
  • 30 cm³ CH₄ reacts with 60 cm³ O₂.
  • Since there's 100 cm³ O₂, O2 is in excess by 40 cm³.
  • CH₄ is the limiting reactant.
• Step 2: Calculate the volume and composition of the gas at the end.
  • CH₄ reacts to produce 30 cm³ CO₂.
  • The total volume of gas at the end is 40 cm³ O2 (left unreacted) + 30 cm³ CO₂ = 70 cm³.

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Excess Calculations - Reacting Gas Volumes