Question

6.4 L/Hr of Isopropyl alochol is fed to a heater with pure Oxygen. the molar flow rate of Oxygen is twice that of the Isopropyl alcohol in the feed. The fraction conversion of O2 is .91. The ratio of CO2 to CO in the effulent stream is 15.5 to 2. Find the molar flow rate of all products in the effulent stream. Please show all work and draw a flow diagram for visual. Thanks

Answer 1

Ans. Density of isopropyl alcohol = 0.7785 g/ml

given volumetric flow rate V=6.4l/hr

molecular weght of isopropanol alcohol MW = 60 g/mol

molar flow rate of isopropanol alcohol is M₁ =( x V)/MW = (0.7785*6400)/(60*3600) = 0.023mol/sec

molar flow rate of oxygen is M₂ = 2 xM₁ =2*0.023 =0.0461 mol/s

2C₃H₇OH + 9O₂    6CO₂ + 8H₂O

at t = 0 (0.023) (0.0461) (0) (0)

at t=t sec fraction of conversion = moles converted / moles fed

fractional conversion of O₂ = 0.91

amount converted or reacted = final number of moles - initial no.of moles = 0.91*0.0461 =0.042 mol/s

moles left = 0.042-0.0461 = 0.0041 mol/s.

since 2 moles of isopropyl a;cohol requires 9 moles of oxygen.

0.042 moles of oxygen requires x no,of moles of isopropyl alcohol.

x = (0.042*2)/9 =0.00933 moles/s of isopropyl reacted

similarly the molar flow rate of products

CO₂ =( 0.00933*.6)/2 =0.028 mol/s

CO = (2*CO₂)/15.2 = 0.00368 mol/s

H₂O = (0.00933*8)/2 = 0.03732 mol/s