Question

Ethylene oxide is produced by the catalytic oxidation of ethylene. The feed into the reactor contains 1700 g/min ethylene and 3700 moles/min air. 2C2H4+O2 --> 2C2H4O

a) What is the limiting reactant?

b) What is the extent of reaction if only 81% of the limiting reactant is converted to the product?

c) What is the molar composition of the product stream?

Answer 1

C2H4 + O2 --> 2CH4O

2C2H4 + O2 --> 2C2H4O

1700 g/min Ethylene

3700 moles of air

MW of Ethylene = 28 g/gmol

m,ol of Ethy = 1700/28 = 60.7 mol of Ethyox

Find mols of O2 in air

3700*(0.21 mol of O2 per mol of air ) =777 mol of O2

therefore, th elimiting reactant is Ethylene Oxide.

b)

Extent of reaction if only 81% of limiting reactant is converted

The extent of reaction: 0.81 mol * 60.7 mol of Eth = 49.17 mol of Ethylene

c)

]Molar composition os product:

N2, O2, C2H4O, C2H4

Balances:

N2 in = N2 out

3700*.79 = 2923 mol of N2

O2 balance:

O2 in - O2 reat = Oout

777 - 0.81*49.17 /2 = 57.08 mol of O2 out

C2H4 balance

C2H4 in - C2H4 reacted = C2H4 out

60.7 -49.17 = 11.53 mol of C2H4 out

C2H4O blaance

C2H4O

in - produced = out

0 - 49.17 = out

49.17 mol C2H4O are produced

Total mols:

2923 + 57.08 +11.53 +49.17 = 3040.8

Fractions:

N2 =2923 /3040.8 = 0.961

O2 = 57.08 /3040.8 = 0.019

C2H4 =11.53 /3040.8 = 0.004

C2H4O = 49.17 /3040.8 = 0.016