Question

The volume of air required to burn 20 ml of methane if air contains 20% oxygen is




koi to daya karke boldo​

Answer 1

For methane:

CH4 + 2O2 = CO2 + 2H2O

At STP, 1 mL of methane requires 2 mL of oxygen. (On applying Gay-Lussac's law).

So, 10 mL of methane will require 20 mL of oxygen.

Air contains 20% oxygen by volume.

Volume of air needed for 20 mL of oxygen = 20 x 100/20 = 100 mL.

2. For acetylene:

2C2H2 + 5O2 = 4CO2 + 2H2O

At STP, 2 mL of acetylene requires 5 mL of oxygen.

So, 10 mL of acetylene will require 25 mL of oxygen.

Volume of air needed for 25 mL of oxygen = 25 x 100/20 = 125 mL.

Total volume of air needed = 100 + 125 = 225 mL