Question

2C2H4+O2 => 2C2H4O

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

a. What is the limiting reactant?

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

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

Show all work or no points will be awarded.

Answer 1

2C2H4(g) + O2(g) → 2C2H4O(g)
twice as many moles of C2H4 are needed to react with O2

1700 g/min ethylene means 1700 / (molecular mass of ethylene) = 1700 / 28.05 = 60.60 moles/minute

3700 moles/min = 3700 / (molecular mass of O2) = 3700/16 = 231.25

They want 60.6/2 mole of O2 for 60.6 mole ethylene
they added an excess of O2
a) So C2H4 is your limiting reagent

b) 81% of limiting reagent means 60.60 * 81 / 100 = 49.08 moles

while reacting 49.08 mole we get = 49.08 C2H4O

since efficiency is only 81%, To get 49.08 C2H4O we have to provide 60.60 mole C2H4 and 30.3 mole O2

The extent of reaction is that cumulative fraction of the reactants that react after a given time, from some reference starting point in time (that is, from a defined time t=0).

In first one minute we will give 60.60 mole C2H4 but 49.08 mole will react and 11.52 mole C2H4 will left over so to complete this we have to need 11.52/60.60 minutes = 11.4 second so total we need 1 minute 11.4 second

c) If we are giving

since efficiency is only 81%, To get 49.08 C2H4O we have to provide 60.60 mole C2H4 and 30.3 mole O2

so in the product stream there will be 49.08 mole C2H4O , (60.60 mole - 49.08 mole) C2H4 (30.3 mole - 24.54mole) O2

SO MOLAR COMPOSITIONS OF PRODUCT IS 49.08 C2H4O  and 11.52 C2H4  + 5.76O2