Question

what amount of oxygen
Required to complete combustion of
of 2 Litre Butan gas.
C4H10
main componet of LP.G.​

Answer 1

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.

Answer 2

Answer:

2 L of butane requires 13 L of oxygen to combust.

Explanation:

Combustion of Butane gives us Carbon dioxide and water

Balanced equation is :

2C4H10 + 13O2 ---> 8CO2 + 10H2O

According to balanced equation :

2 moles of Butane gas on combust with 13 moles of oxygen to give 8 moles of CO2

As we know that 1 mole of any gas  at STP occupy 22.4 L

so 2 moles of butane will occupy 2x22.4L = 44.8 L

13 moles of oxygen will occupy = 13x 22.4L=291.2 L

2moles of butane [ 44.8L] will combust with 291.2 L of oxygen

2 L of butane requires --------? oxygen

=291.2x2/ 44.8

=582.4/44.8

=13L

∴ 2 L of butane requires 13 L of oxygen to combust.