In the production of formaldehyde (CH_2O) by catalytic oxidation of methanol (CH

Question

In the production of formaldehyde (CH_2O) by catalytic oxidation of methanol (CH_3OH) an equimolar mixture of methanol and air (21% oxygen and 79% nitrogen) is sent to a catalytic reactor. The reaction is catalyzed by finely divided silver supported on alumina as suggested in Figure 7.1 where we have indicated that carbon dioxide (CO_2) is produced as an undesirable product. The conversion for methanol (CH_3OH) is given by C = Conversion of CH_3OH = - CH_3OH/(M_CH_3OH)_1 = and the selectivity for formaldehyde/carbon dioxide is given by S = Selectivity of CH_2O/CO_2 = CH_2O/CO_2 = 8.5 In this problem you are asked to determine the mole fraction of all components in the Stream #2 leaving the reactor.

Explanation / Answer

The reactions are

CH3OH + ½ O2 HCHO + H20 (1)

CH3OH + 1.5O2----àCO2+2H2O (2)

Conversion = moles of CH3OH reacted/ moles of CH3OH fed= 0.2

Selectivity = moles of COOH formed/moles of CO2 formed= 8.5

Let 1 mole of CH3OH and 1 mole of air are fed

Moles of oxygen fed =0.21, moles of methanol fed= 1 mol

Moles of methanol reacted= 0.2 mole

Let x= moles of methanol reacted through reaction -1. Moles of COOH formed = x

0.2-x= moles of methanol reacted through reaction-2. Moles of CO2formed = 0.2-x

Hence x/(0.2-x)= 8.5, x= 8.5*0.2-8.5x, 9.5x= 8.5*0.2, x= 0.179, moles of COOH formed= 0.179, moles of CO2 formed =0.2-0.179= 0.021 .

Moles of O2 consumed = 0.5*0.179+1.5*0.021=0.121 moles ( Oxygen consumed from reaction-1 and oxygen consumed for reaction-2)

Products : COOH : 0.179, CO2= 0.021, N2= 0.79 ( From air ), moles of O2 remaining = 0.21-0.121= 0.089, moles of water= 0.179+2*0.021= 0.221, CH3OH=1-0.2=0.8

Total moles of products : 0.179+0.021+0.79+0.089+0.221= 1.3

Composition ( not considering unreacted CH3OH) : COOH= 0.179/1.3= 0.14, CO2= 0.021/1.3=0.016, N2=0.79/1.3= 0.61, O2= 0.089/1.3= 0.068, H2O= 0.221/1.3= 0.17,

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