What volume of air will be required to burn completely?

The volume of air required to completely burn 1L of CO is 2.5 litre air. Hence, the correct option is (B).

What is the volume of air required for complete combustion of 20 Litre of methane?

=200L of air.

What volume of air is 21?

3200L
If O2 is 672 L, then air =10021×672L=3200L.

What is the volume of air required for the complete combustion of 20 Litre of methane air contains 20\% by volume O2?

22.4L if CH4 requires 44.8L of oxygen.

What volume of oxygen is required for complete combustion of?

Grade 9. Thus for 1 mole methane we need 2 mole of oxygen for complete combustion. Hence 0.5 moles of oxygen will occupy 22.4 L x 0.5 = 11.2 L. 11.2 L of oxygen is required for complete combustion of 0.25 mole methane at STP.

How do you calculate the volume of air for complete combustion?

∴ Total volume of actual products of combustion = (V1 + V) m3. Let O be the percentage of oxygen (by volume) present in the exhaust gases, then the excess quantity of air which will contain this volume of oxygen will be O/21 m3.

What volume of air is required to burn 1cm^3 of methane?

The volume of air required to burn Methane completely is 100cm^3.And the volume of air required to burn acetyle 1 volume of methane requires 2 volumes of oxygen to burn in it completely. So, 1cm^3 of methane requires 2cm^3 of Oxygen to burn completely.

How do you calculate volume of air needed for acetylene gas?

Volume of air needed for 20 mL of oxygen = 20 x 100/20 = 100 mL. 2. For acetylene: 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.

How much oxygen is required to produce 10 cm3 of C2?

As air contains 20\% oxygen, so 100 cm 3 of air is needed. needs 2.5 volume of oxygen. So, 10 cm3 of C 2 needs 25 cm3 of oxygen. As air contains 20\% oxygen, so 125 cm3 of air is required containing 25 cm3 of oxygen. Was this answer helpful?

How much air is required to burn 21\% of oxygen?

Assuming 21\% is volume of oxygen in air, we would need 27.43/1,430/0.21 M^3 or 0,091 M^3 (or CM) of air stoichiometrically. In practice, it is impossible to get the fuel burnt with exact supply of stoichiometric air. Depending on the burner efficiency and the process, you might have to give excess air ranging from 10\% (minimum requirement) to 50\%