Production of formaldehyde (CH,O) incots te cndl oxidation of methanol (CHOH) Th

Question

Production of formaldehyde (CH,O) incots te cndl oxidation of methanol (CHOH) The reaction is accompanied by a side reaction where some of the peoduct es oxygen to produce CO and H:O 2 The feed to the reactor contains designed to yield 70% conversion of methanol and the product leaving the rea expected to have the selectivity of 6 kmol CH:Okmol of formaldehyde, determine the amount of composition (%) ofoutlet gas in dry basis. mdr is methanol and 100% excess air. Suppose CO. For the production of 900 kphe ine the amount of methanol required ghr) and the mar (40 Marks)

Explanation / Answer

The production rate is 900 kg/hr

The production rate in mol/hr = 900 kg/hr / (30 g/mol) == 30 kmol/hr

The moles of CO will be produced = 30 kmol/hr CH2Ox (1 kmol CO / 6 kmol CH2O)

= 5 kmol/hr CO

So moles of CH3OH required from reaction-2 = 5 kmol/hr CO x( 1 kmol/hr CH3OH / 1 kmol/hr CO ) = 5 kmol/hr CH3OH

moles of CH3OH required from reaction-1 = 30 kmol/hr CH2O x ( 1 kmol/hr CH3OH / 1 kmol/hr CH2O ) X (1/0.7)

= 42.85 kmol/hr

Total CH3OH required = 42.85 kmol/hr + 5 kmol/hr = 47.85 kmol/hr

So, mass rate of CH3OH = 47.85 kmol/hr x 32 gg/kmol == 1531.2 kg/hr

Moles rate of CO formed = 5 kmol/hr

Moles rate of CH2O formed = 30 kmol/hr

Moles of O2 reacted = 0.5 x 5 kmol/hr + 0.5 x 30 kmol/hr == 17.5 kmol/hr

Moles of O2 left = 30 kmol/hr - 17.5 kmol/hr = 12.5 kmol/hr

Moles of CH3OH left == (47.85-30 ) kmol/hr == 17.85 kmol/hr

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