Methanol is being produced according to the reaction CO+ 2H CH, OH Asynthesis ga
ID: 533213 • Letter: M
Question
Methanol is being produced according to the reaction CO+ 2H CH, OH Asynthesis gas stream consisting of 66.20% Hz, 33 40% co, and o400% CH4 (by mole) is fed into the process described below at a rate of 100 molhour. The effluent stream from the reactor is fed into a and the reactor. To avoid the buildup of methane, a portion of the stream is purged. At present operating conditions, the CO Conversion over the entire process is 90 and the methane mole fraction at the reactor inlet (Stream 2) is 2.870%. Reactor Separator methanol purge What is the production rate of methanol? Number mol CH, OH hrExplanation / Answer
Ans:- The given reaction is
CO+2H2 -> CH3OH
Now 66.2% H2 means 66.2g of H2
Therefore number of moles of H2 = 66.2 /2
= 33.1
Again 33.4%CO means 33.4g of CO
Therefore number of moles of CO = 33.4 / 28
= 1.192
0.4% CH3OH means 0.4 g CH3OH
Therefore number of moles of CH3OH = 0.4/32
= 0.0125
Now , mole fraction of H2 = 33.1 / 33.1+1.192+0.0125
= 0.964 mol H2 /mol
mole fraction of CO = 1.192 / 33.1+1.192+0.0125
= 0.034 mol CO/mol
mole fraction of CH3OH = 0.0125/ 33.1+1.192+0.0125
= 0.000364 mol CH3OH/mol
Here total molar flow rate = 100mol/hr
100 = purge flow+ reactor flow
Again mole fraction of methanol = purge flow + 2.87 reactor flow
0.000364= purge flow + 2.87 reactor flow
From both the reaction
we get 100-0.000364=2.87Reactor flow - reactor flow
Reactor flow =53.45
Production flow rate = 100- 53.45
= 46.55mol/min
Molar flow rate = 33.1+1.192+0.0125/100
= 0.3430 mol/hr
Ratio = 100/0.3430