Question #230837

A petrochemical plant in Jurong Island produces formaldehyde (CH2O) industrially by the catalytic oxidation of methanol (CH3OH). The following is the chemical reaction: CH3OH + Â½O2 â†’ CH2O + H2O --- (Equation 1) Unfortunately, a significant portion of the formaldehyde will react with oxygen to produce CO and H2O. The following is the side reaction: CH2O + Â½O2 â†’ CO + H2O --- (Equation 2) Assume twice the stoichiometric amount of air needed for oxidation (include Equation (1) and Equation (2)) is fed to the reactor. The conversion of methanol is 90% in the reactor. The outlet stream of the reactor was analysed and found to be 150 kg/hr of formaldehyde, 30 kg/hr of CO and other components.

(i) Determine the mass flowrate (kg/hr) of Stream A. ( 4 marks ) (ii) Determine the molar flowrate (kg-moles/hr) of Stream B. ( 4 marks ) (iii) Determine the mass flowrate (kg/hr) of water in the outlet stream. ( 4 marks ) (iv) Determine the molar composition in the outlet stream. ( 5 marks )

Expert's answer

"CH_3OH + \u00bdO_2 \u2192 CH_2O + H_2O --- (Equation \\space 1)\\\\\n\\implies x \\space mole \\space \\space \\space \\space \\space 2x \\space mole \\space \\space \\space \\space \\space 0 \\space \\space \\space \\space \\space 0\\\\\n\\implies x(1-0.9) \\space mole \\space \\space \\space \\space \\space 2x-\\frac{0.9x}{2} \\space mole \\space \\space \\space \\space \\space 0.9x \\space \\space \\space \\space \\space 0.9x\\\\\nCH_2O + \u00bdO_2 \u2192 CO + H_2O --- (Equation \\space 2)\\\\\n\\implies 0.9x \\space \\space \\space \\space \\space \\space 2-0.9x \\space \\space \\space \\space \\space \\space 0 \\space \\space \\space \\space \\space 0\\\\\n\\implies 0.9x-y \\space \\space \\space \\space \\space \\space 2x-\\frac{0.9x}{2} \\space \\space \\space \\space \\space \\space y \\space \\space \\space \\space \\space y\\\\"

Mole of formaldehyde in the outlet

"\\frac{150 kg\/hr}{30g\/mol}= 5000 mol\/hr\\\\\n0.9x-y=5000\\\\\ny = \\frac{30 kg\/hr}{28g\/mol}=1071.428 mol\/hr\\\\\n0.9x=500+1071.428\\\\\nx=6746.03 mol\/hr"

Part (i)

To find theÂ the mass flow-rate (kg/hr) of Stream A we can denote it as x

"x= 6746.03 \\frac{mol}{hr}\\\\\nx= 6746.03 \\frac{mol}{hr}*\\frac{32 g}{mol}*\\frac{1 kg}{1000g}\\\\\nx= 215.87 kg\/hr"

Part (ii)

To determine the molar flow rate (kg-moles/hr) of Stream B we can denote it as y

"2x- \\frac{0.9 y}{2}- \\frac{z}{2}= Air \\space left\\\\\n2x- \\frac{0.9*6746.03}{2}- \\frac{1071.42}{2}= 9920.63\\\\\nSteam \\space B = 2y = 2*13492.06 mol\/hr\\\\\n\\implies y= 13.49 kgmol\/hr"

Part (iii)

Mass flow rate of water = "0.9x+y= 7142015 mol\/hr"

Mass flow rate = "128.57 kg\/hr"

Mass flow rate of water = "0.9x+y= 7142015 mol\/hr"

Part (iv)

The molar composition of CO = "\\frac{1071.428}{23809.5119}*100=4.5\\%"

The molar composition of H_{2}O = "\\frac{7142.85}{23809.511}*100=30\\%"

The molar composition of CH_{3}OH = "\\frac{674.603}{23809.511}*100=2.84\\%"

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