Answer to Question #230837 in Chemical Engineering for mya

Question

A petrochemical plant in Jurong Island produces formaldehyde (CH2O) industrially by the catalytic oxidation of methanol (CH3OH). The following is the chemical reaction: CH3OH + ½O2 → 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 + ½O2 → 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 )

Answer 1

$CH_3OH + ½O_2 → CH_2O + H_2O --- (Equation \space 1)\\ \implies x \space mole \space \space \space \space \space 2x \space mole \space \space \space \space \space 0 \space \space \space \space \space 0\\ \implies x(1-0.9) \space mole \space \space \space \space \space 2x-\frac{0.9x}{2} \space mole \space \space \space \space \space 0.9x \space \space \space \space \space 0.9x\\ CH_2O + ½O_2 → CO + H_2O --- (Equation \space 2)\\ \implies 0.9x \space \space \space \space \space \space 2-0.9x \space \space \space \space \space \space 0 \space \space \space \space \space 0\\ \implies 0.9x-y \space \space \space \space \space \space 2x-\frac{0.9x}{2} \space \space \space \space \space \space y \space \space \space \space \space y\\$

Mole of formaldehyde in the outlet

$\frac{150 kg/hr}{30g/mol}= 5000 mol/hr\\ 0.9x-y=5000\\ y = \frac{30 kg/hr}{28g/mol}=1071.428 mol/hr\\ 0.9x=500+1071.428\\ x=6746.03 mol/hr$

Part (i)

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

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

Part (ii)

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

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

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 H₂O = $\frac{7142.85}{23809.511}*100=30\%$

The molar composition of CH₃OH = $\frac{674.603}{23809.511}*100=2.84\%$