Ethane of volume 400 cm ^3 was completely burnt in excess oxygen according to the following equation: 2C 2 H 6…

Analyze the question: This is a gas volume problem. By Gay-Lussac's law, gases measured at the same temperature and pressure react in volumes that are in the same ratio as their mole ratio in the balanced equation. Steam (H₂O gas) counts as a gas here.

Concept: The coefficients in the equation give the volume ratio directly:

2C₂H₆ : 6H₂O 2 volumes of ethane produce 6 volumes of steam, so the ratio is 1 : 3.

Solution:

• 2 volumes of C₂H₆ give 6 volumes of H₂O
• 400 cm³ of C₂H₆ gives x cm³ of H₂O

x = (6 / 2) × 400 = 3 × 400 = 1200 cm³

Check: 400 cm³ is 200 × 2, so the volume scale factor is 200. Steam is 6 × 200 = 1200 cm³.

Tip: For gas volume questions, don't convert to moles. Use the coefficients as volume ratios, and write "given volume × (wanted coefficient ÷ given coefficient)". Always check that the gases are at the same temperature and pressure. Here the oxygen is in excess, so ethane is the limiting reactant and sets the amount of steam.