C 2 H 4(g) + 3O 2(g) → 2CO 2(g) + 2H 2 O (g) The above equation represents the combustion of ethene.If 10cm ^3…

Analysis

To solve this problem, we use Gay-Lussac’s Law of Combining Volumes, which states that when gases react, they do so in volumes which are simple ratios to one another and to the products, provided temperature and pressure remain constant.

Concept

The balanced chemical equation tells us the molar ratio of the reactants: C$_2$H$_{4(g)}$ + 3O$_{2(g)}$ → 2CO$_{2(g)}$ + 2H$_2$O$_{(g)}$

From the equation:

• 1 mole (or 1 volume) of C$_2$H$_4$ reacts with 3 moles (or 3 volumes) of O$_2$.

Step-by-Step Solution

1. Determine the required volume of oxygen: Since 1 volume of C$_2$H$_4$ requires 3 volumes of O$_2$: 10 cm$^3$ of C$_2$H$_4$ will require: $10 \text{ cm}^3 \times 3 = 30 \text{ cm}^3$ of O$_2$.

2. Calculate the remaining volume of oxygen: We were provided with 50 cm$^3$ of oxygen at the start. Volume of oxygen used = 30 cm$^3$. Volume of oxygen remaining = (Initial volume) - (Volume used) Volume of oxygen remaining = $50 \text{ cm}^3 - 30 \text{ cm}^3 = 20 \text{ cm}^3$.

Conclusion

After the reaction is complete, 20 cm$^3$ of oxygen remains unreacted. This matches option A.

Exam Tip

Always check the stoichiometry of the balanced equation first! A common mistake is to assume a 1:1 ratio. In combustion reactions of hydrocarbons, the oxygen requirement is always determined by the coefficients in the balanced equation. If you are ever unsure, quickly re-balance the equation to ensure the number of atoms on both sides is equal before calculating the volumes.