Answer: The enthalpy of formation of is -396 kJ/mol
Explanation:
Calculating the enthalpy of formation of
The chemical equation for the combustion of propane follows:
The equation for the enthalpy change of the above reaction is:
We are given:
Putting values in above equation, we get:
The enthalpy of formation of is -396 kJ/mol
The number of mole will be 65.81 mole.
An ideal gas would be one for which both the overall volume of the molecules and even the forces that exist between them are so negligible as to have no influence on the behavior of something like the gas.
Number of ideal gas can be calculated by using the formula:
PV = nRT
where, p is pressure, n is number of mole, R is gas constant and T is temperature.
Given data:
V= 1750 = 1750 L
P = 125,000 p = 1.2 atm
R = 0.082 L /mole kelvin
T = 273+127 = 400 K
Now, put the value of given data in above equation.
1.23atm x 1750L = n x 0.0820atm x Liter/ mole x kelvin x 400K
n = 65.81 mole.
Therefore, the number of mole will be 65.81 mole
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