Question

PROBLEM 10 Methyl iodine can be produced by reacting hydriodic acid with an excess of methanol via the reaction HI+CH,OHCH,I+H2O A typical process for large-scale production of methyl iodine is indicated in the figure. The process conditions are: The fresh feed contains 1 mole of HI (stream #1] (Calculating Basis) Fresh methanol (stream #3) is fed 300% in excess with respect to the fresh feed of HI. 3. The overall conversion of Hl is 70% The stream (#7) exiting the second separator contains 10% water. 5. 25% of the Hl in stream #4, return to the reactor with the recycle (stream #10). 6. All compositions are on a molar basis. CHOH CHOR H.O Reactor a. What is the single-pass conversion of HI and that of methanol What is the overall conversion of methanol b. Calculate the excess of methanol in the reactor, with respect to stream 22 C. Determine all quantities and summarize your findings in the following table. NOTE: CLEARLY, WRITE DOWN THE CALCULATION PROCEDURE

Answer 1

Solution

CH2ol1 If 8( fel to (fig) HI (fa) Heo HI + C 014 - CH₂ I q + H₂O bet fi, F2- ifio mall see be the Mola flow roses of Boria Imole HJ Siuve Chyor us 3004. Excess XI . & Theis F3 = 300 - Moles of CholH fead Mores of CH₂ on regu. . Modes of CH₃OH requi xloo * 200= Moles fed 1 x 100 Moles feale 4 and (See = fal

dinining reagent cat U Overall Conversion 3 Holes of Cazt obtained out of the provers out of the Moles of 19 fed in Stream I 7 70 Moles of CH₂9 obtained ow in fo) Xhno - Moles of CHgil obtained in for 0.7 moll see Thuis Moles of CH₂ I in stream fy = 0.7 and 1sec Mecause since only rg and Hio are exiting from a thus there Must Nut we ary CH39 in Strean 8 as well. Tum in stream y Moles of H20 = Moles of Chad produces produced = 0. Tuoll see Там H3 + CHрін — оң 3 + H2O at 1-0 1. Sp. S 0 0 at t-os osoit 4-6.7 0.7 0.7 ?-03 - 3.3 Thus, Composition of stream 4a H9 = 1.3 malis CH₂014 a 3.3 molls Ct₂ I = 0.7 molls H2O = 07 udlsec

Now Lince CH₂ ore se exiting fresor Stream 7 Thus, oles of Ctzole in Stream - 3.3 md I see ol of water in t'a sheam 0.1 = Molou hare of H₂O Total Molas rate A 0:12+ 0.33 =2 - X 0.33 0.90 = 0.3667 rud/sec a od H₂O Melou hare of H₂0 = x In stream 7 Molor flow rate of HI in stream to = 25 x 0.3 100 Titite 4 = 0.075 dll see HI in stream 9 = (0.3-0.075) = 0.225 roll see Thu HI Entering the leason 0.075 + 1 1.045 Mol I see it This single pars Conversion = HI reaver H9 few.is reactos reaved x too

Since HI dearen- CH₂J formed - 07 model Sec The single pour conversion 0.7yluo 14076 65.116 % / (Single pose Conversion = Overall Conversion of Meehand - y too Methanal reacted Methan of fedin Stream 3 xloo = o.f 17-5-4. Conversion of CHzoll (b) Excers of Methana CHzoll = (113011 in streamz) . . - Chiz dil required Stoichiomere cally with respect to HBO CH₃OH required Stoich's autorically wert. HI in stream X loo 4- (1.075) x Xloo 1.075 2.925 xlov - 272.09 % Excene Mettiano 1075