What amount of oxygenRequired to complete combustion ofof 2 Litre Buta

1.	What amount of oxygenRequired to complete combustion ofof 2 Litre Butan gas. C4H10 main componet of LP.G.
Answer» combustion reaction of butane is .. $2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_20$ here it is CLEAR that, for 2 mole of butane,13 mole of oxygen is REQUIRED to COMPLETE combustion reaction. given, VOLUME of butane = 2L so, mole of butane = 2L/22.4L = 1/11.2 now, for 2 moles of butane require 13 moles of oxygen, so, 1/11.2 mole of butane require 13/2 × 1/11.2 = 13/22.4 mol of oxygen. so, amount of oxygen = number of mole of oxygen × 22.4 L = 13/22.4 × 22.4 L = 13L hence, 13L oxygen is required to compete combustion of 2L butane.

What amount of oxygenRequired to complete combustion ofof 2 Litre Butan gas. C4H10 main componet of LP.G.

Answer»

combustion reaction of butane is ..

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_20$

here it is CLEAR that, for 2 mole of butane,13 mole of oxygen is REQUIRED to COMPLETE combustion reaction.

given, VOLUME of butane = 2L

so, mole of butane = 2L/22.4L = 1/11.2

now, for 2 moles of butane require 13 moles of oxygen,

so, 1/11.2 mole of butane require 13/2 × 1/11.2 = 13/22.4 mol of oxygen.

so, amount of oxygen = number of mole of oxygen × 22.4 L

= 13/22.4 × 22.4 L

= 13L

hence, 13L oxygen is required to compete combustion of 2L butane.