Asked by Anna
A 2.00 L flask is filled with propane gas (C3H8)at 1.00 atm and -15.0 Degrees Celsius. What is the mass of the propane in the flask?
Answers
Answered by
DrRebel
1.35gms.
From the Ideal Gas Law PV=nRT solve for moles CH4 = n = PV/RT = (1-Atm)(2.00-L)/(0.08206-L-Atm/mol-K)(273+15)K = 0.085-mole CH4... Convert to grams by multiplying by formula wt of CH4 = (0.085-mol)(16-gms/mol) = 1.35 gms.
From the Ideal Gas Law PV=nRT solve for moles CH4 = n = PV/RT = (1-Atm)(2.00-L)/(0.08206-L-Atm/mol-K)(273+15)K = 0.085-mole CH4... Convert to grams by multiplying by formula wt of CH4 = (0.085-mol)(16-gms/mol) = 1.35 gms.
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