Question

4. 1.56 Kmols/hr of HI is reacted with CH3OH to produce CH3I according to the following reaction

HI + CH3OH =CH3I + H2O

HI is separated from the product stream and recycles back into the reactor. The product contains 50 mol % CH3I and 50 mol % CH3OH and the waste stream contains 40 mole % HI and 60 mole % H2O. The single pass fractional conversion of the process is 40%. Calculate: Methanol added & Amount of HI recycles?

Answer 1

HI Balance:

Let the amount of HI which have reacted = x

input – consumption = output

- x = 0.4 W ............. (1)

H2O balance:

Input + generation = output

0 + x = 0.6 W ..............(2)

solving (1) and (2) simultaneously gives:

W = 1.56 kgmole/hr X = 0.936 kgmoles/hr.

CH3 I Balance:

Generation = Output

0.936 = 0.5 PP =1.872 kgmole/hr

Methanol Balance:

Input –Consumption = Output

M – 0.936 = 0.5x1.872M = 1.872 kgmole/hr

fractional conversion = 40%

Fractional conversion:

f = HI(rxn) / HI ( reactor fed)

HI fed to the reactor

= 1.56 + R

HI reaction

= 0.936 = 0.4(1.56+R)R = 0.78 kgmole/hr