Question

XP10. Methanol (CH3OH) can be used as an antifreeze or fuel and can be synthesized from carbon monoxide (CO) and hydrogen (H2). The product from a reactor contains CO, H2, CH2OH, and 102 mol/h N2 (an inert component). The total flow rate of the feed to the reactor is 476 mol/h, and the feed contains an equal molar flow rate of CO and H2. The total flow rate exiting the reactor is 318 mol/h. (a) Draw and label a process flow diagram. (b) Calculate the molar flow rates (mol/h) of each component exiting the reactor.

Answer 1

Part a

Flow diagram

Part b

Product contains N2 = 102 mol/h

Feed rate = 476 mol/h

Feed contains CO, H2, N2. (N2 is inert)

Molar flow rate of CO + molar flow rate of H2 = 476 - 102

= 374 mol/h

Feed contains equal molar flow of CO and H2

Molar flow rate of CO = molar flow rate of H2 = 374/2

= 187 mol/h

The balanced reaction

2H2 + CO = CH3OH

From the stoichiometry of the reaction

1 mol H2 required = 0.5 mol CO

187 mol H2 required = 187*0.5 = 93.5 mol CO

Limiting reactant = H2

Excess reactant = CO

Moles of CH3OH produces = 187*0.5 = 93.5 mol

At exit

Total molar flow = 318 mol/h

N2 = 102 mol/h

Molar flow of CH3OH + molar flow of H2 + molar flow of CO

= 318 - 102 = 216 mol/h

0.5*molar flow of H2 reacted + molar flow of H2 at exit + (187 - 0.5*molar flow of H2 reacted) = 216

0.5*molar flow of H2 reacted + (molar flow of H2 at fed - molar flow of H2 reacted) + (187 - 0.5*molar flow of H2 reacted) = 216

0.5*molar flow of H2 reacted + (187 - molar flow of H2 reacted) + (187 - 0.5*molar flow of H2 reacted) = 216

molar flow of H2 reacted = 158 mol/h

Molar flow of H2 at exit = 187 - 158 = 29 mol/h

Molar flow of CO at exit = 187 - 0.5*158 = 108 mol/h

Molar flow of CH3OH at exit = 0.5*158 = 79 mol/h