Show Your Work Use EASC2211 pure ethanol (C_2.H_6 O - 46 g/mol) is to be burned
ID: 518555 • Letter: S
Question
Show Your Work Use EASC2211 pure ethanol (C_2.H_6 O - 46 g/mol) is to be burned with air(20 mol% O_2 and 80 mol% N_2) in a burner to provide heat. The food rate of the propane 920 g/min and 20% excess air is used. If 90% of the propane fed to the burner was consumed, do the following: a) Complete the diagram, defining variables for the unknown feed b)Use appropriate mass balances to determine the rate of reaction. c) What is the actual flow rate of the air? d) Determine the composition of the product gasExplanation / Answer
The balance reaction is C2H6O+ 3O2------->2CO2+3H2O
Rate of mass of C2H6O- rate of loss due to chemical reaction = rate of mass out Since 90% of ethanol sent is combusted,
920 g/min -920*0.9= rate of mass out
Rate of mass out = 920*0.1 =92 g/min
Rate of mass of O2 in- rate of mass of oxygen consumed= rate of oxygen out
Since N2 is not participating in the reaction, rate of N2 in = Rate of N2 out
2. molar flow rate of ethanol= 920 g/min, molar flow rate of ethanol= 920/46= 20 gmoles/min ( moles= mass/molar mass)
as per the stoichiometry of the reaction, moles of O2 required= 3*20= 60 moles/min ( 1 mole of C2H6O requires 3 moles of O2)
Air contains 20%O2 and 80% N2, molar flow rate of air required = 60/0.2= 300 moles/min
molar flow rate of air is 20% excess than stoichiometric requirement, air supplied= 300*1.2= 360 moles/min
Products flow rate ( moles/min): unreated C2H6O= C2H6O supplied- C2H6O reacted= 20-20*0.9 =20*0.1= 2 , O2= O2 supplied- O2 consumed= 360*0.2-18*3=18, CO2= 2* moles of C2H6O reacted = 2*18= 36, moles of H2O= 3* moles of C2H6O consumed= 3*18= 54 moles/min, N2= 360*0.80 =288
total flow rate of products = 2+ 18+36+54+288=398
Composition : C2H6O= 2/398= 0.005, O2= 18/398 = 0.045, CO2= 36/398= 0.090, H2O= 54/398=0.1356 and N2= 288/398=0.7487