The propane tank for my barbeque holds seven pounds of liquid propane (C3H8). a)
ID: 1088706 • Letter: T
Question
The propane tank for my barbeque holds seven pounds of liquid propane (C3H8).
a) If the tank ruptured, what would be the size of the cloud of propane which forms?
b) If the cloud of propane mixed with just the right (stoichiometric) quantity of air, and ignited, the resulting reaction would produce 3 moles C02, 4 moles H20 and would have 20 moles N2 from the original air, for each mole of propane combusted. Assuming a final temperature of 1100°C, what would be the size of the resulting fireball in my backyard?
Explanation / Answer
Answer:
Let the number of moles of propane (C3H8) is the cloud = 1 mol
Propane cloud is mixed with air in a ‘just the right’ stoichiometry; it’s assumed that it means that there just sufficient O2 for complete combustion of propane.
C3H8 + 5 O2 + 20 N2------------> 3CO2 + 4 H2O + 20 N2 - complete combustion
In the reaction mixture, all propane and O2 is converted into CO2 and H2O while N2 remains unaffected.
Now,
Total number of moles in the product =3 mol of CO2 + 4 mol of H2O + 20 mol of N2
= 27 mol
Two more assumptions-
I. The gases behave ideally
II. Since the combustion occurs in open air, the pressure of the gas is assumed to be equal to 1 atm.
Given, final temperature of the products = 11000C = 1373 K
Now, Using Ideal gas equation: PV = nRT - equation 1
Where, P = pressure in atm
V = volume in L
n = number of moles
R = universal gas constant= 0.0821 atm L mol-1K-1
T = absolute temperature (in K)
1 atm x V = 1 mol x (0.0821 atm L mol-1K-1) x 1373 K
Or, V = 3041.952 L
Hence, the volume of resultant fireball = 3041.952 L
= 3.04 m3 ; [1 m3 = 1000.0 L]
Note: Since the number of moles of propane is not defined, the size of resultant fireball increases (directly proportional) with increase in initial moles of propane.
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