Solve the following numerical problem. Ethane burns in oxygen according to the chemical equation: 2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O If 80 ml of ethane is burnt in 300 ml of oxygen, find the composition of the resultant gaseous mixture when measured at room temperature.

All gases at the same temperature and pressure, so volumes follow the coefficients.

Step 1 — find the limiting reactant. C₂H₆ : O₂ = 2 : 7 80 ml of C₂H₆ needs (7 ÷ 2) × 80 = 280 ml of O₂ 300 ml is available, so ethane is the limiting reactant.

Step 2 — oxygen left over. unused O₂ = 300 − 280 = 20 ml

Step 3 — carbon dioxide formed. C₂H₆ : CO₂ = 2 : 4 = 1 : 2 CO₂ = 2 × 80 = 160 ml

Step 4 — the water. At room temperature the water formed is a LIQUID, so it contributes no gas volume.

Resultant gaseous mixture: 160 ml of carbon dioxide and 20 ml of unused oxygen (total 180 ml).

"Composition of the resultant gaseous mixture" means list every gas still present at the end and its volume — so both the product and the leftover reactant.

Two marks are routinely dropped here. The first is forgetting the unused oxygen and answering "160 ml of CO₂" alone. The second is including the water. The phrase "measured at room temperature" is in the question precisely to tell you the steam has condensed — water is a liquid at room temperature and occupies no measurable gas volume. Had the question said the volumes were measured at 100 °C or above, the 240 ml of water vapour would have counted.