Question

40 cm³ of methane (CH₄) is reacted with 60 cm³ of oxygen.

The equation for the reaction is given below:

\[\ce{CH4 + 2O2 -> CO2 + 2H2O}\]

All volumes are measured at room temperature.

What is the total volume of the gases remaining at the end of the reaction?

पर्याय

• 60 cm³

• 40 cm³

• 45 cm³

• 50 cm³

Answer 1

40 cm³

Explanation:

\[\ce{CH4 + 2O2 -> CO2 + 2H2O}\]

According to Gay Lussac’s Law, one volume of methane combines with two volumes of oxygen to produce one volume of carbon dioxide. As water (H₂O) is a liquid, it does not contribute to the overall volume of gases.

Therefore, 40 cm³ of CH₄ would require 2 × 40 cm³ = 80 cm³ of O₂. 

Since only 60 cm³ of O₂ is available, 60 cm³ of O₂ reacts with `1/2 xx 60` cm³ = 30 cm³ of CH₄ is used. Now, the volume of remaining CH₄ will be 40 cm³ − 30 cm³ = 10 cm³.

According to equation, 60 cm³ of O₂ will produce `1/2 xx 60` cm³ = 30 cm³ of CO₂.

Therefore, the total volume of gases remaining is the sum of the volume of CH₄ remaining and the volume of CO₂ produced.

So, total volume = 10 cm³ + 30 cm³ = 40 cm³.