Question

based on gay lussac's law
butane C4 h10 burns in air to form water and carbon dioxide calculate the volume of air required to completely 5liters of butane​

Answer 1

Balanced equation:

2 C4H10 + 13 O2 = 8 CO2 +10 H20

Now we have to find the volume of O2 required . Let's go pointwise

•First convert stochiometric coefficients into stochiometric volumes as we have not given any condition.

•Clearly we have given 5L of butane and stochiometry shows we have 2L of butane

• So we have by stochiometry 13L of oxygen

USING UNITARY METHOD

LET the amount of gas required be X.

Now 2x=13×5

x=65/2

x=32.5L

Hence volume of gas required is 32.5 litres

Hope it helps

_____∆_______∆__________∆___