Question

what the volume of Oxygen gas measured at 0 * c and 1 atm is needed to burn completely 1l of prpane gas c3h8 under same condition?

Answer 1

we can deduce the mol of propane so mol(propane) = x

according to the equation (which is balanced)

5 O2 + C3H8 ----> 3 CO2 + 4 H2O

the mol of O2will be: mol(propane) multiplied by 5

thus if we were to plug in mol of O2 int othe forumla

pV = nRT

where p is 101.325 kpa
n is the mol of oxygen
R is 8.31 (STANDARD VALUE)
T is 273 K
with V (the volume) being the unknown
we can deduce the answer

Answer 2

Hey Aman,

● Answer -

Oxygen required = 5 L

◆ Explaination -

Combustion of propane in presence of oxygen takes place as follows -

C3H8 + 5O2 --> 3CO2 + 4H2O

Here, 5 mol of O2 are required to burn 1 mol of C3H8.

Avogadro's law states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.

Hence, 5 L oxygen is required to burn 1 L propane.

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