Consider the following reaction equation C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l). The volume of oxygen at s.t.p…

Analyze the question: We need the volume of O₂ at s.t.p. needed to completely burn 14 g of ethene (C₂H₄), using the balanced equation.

Concept: Use the mole ratio from the balanced equation, together with:

• moles = mass ÷ molar mass
• 1 mole of any gas at s.t.p. occupies 22.4 dm³

Solution:

Balanced equation: C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

Step 1: Moles of ethene n = 14 ÷ 28 = 0.5 mol

Step 2: Mole ratio from the equation 1 mol C₂H₄ reacts with 3 mol O₂, so 0.5 mol C₂H₄ needs 0.5 × 3 = 1.5 mol O₂

Step 3: Convert to volume at s.t.p. V = 1.5 × 22.4 = 33.6 dm³

Tip: In gas-volume questions, follow this chain: mass → moles → mole ratio → volume. Here, 14 g is exactly half of the molar mass (28 g), so you can quickly see that you need half of the volume for 1 mole of ethene: 3 × 22.4 = 67.2, and half of that is 33.6 dm³.