Acetylene is hydrogenated to form ethane. The feed to the reactor contains 1.50 mol H₂/mol C₂H₂.

(a) Calculate the stoichiometric reactant ratio (mol H₂ react/mol C₂H₂ react) and the yield ratio (kmol C₂H₆ formed/kmol H₂ react).

(b) Determine the limiting reactant and calculate the percentage by which the other reactant is in excess.

(c) Calculate the mass feed rate of hydrogen (kg/s) required to produce 4×10⁶ metric tons of ethane per year, assuming that the reaction goes to completion and that the process operates for 24 hours a day, 300 days a year.

(d) There is a definite drawback to running with one reactant in excess rather than feeding the reactants in stoichiometric proportion. What is it?

[Hint: In the process of Part c, what does the reactor effluent consist of and what will probably have to be done before the product ethane can be sold or used?]

Answer:

We have the balanced reaction as follows:

C₂H₂ + 2H₂  → C₂H₆

(a)

Stoichiometric ratio mol H₂ react/mol C₂H₂ react = 2 mol H₂ / 1 mol C₂H₂
yield ratio kmol C₂H₆formed/kmol H₂ react   = 1 kmol C₂H₆ / 2 kmol H₂

(b)

we have   1.50 mol H₂/mol C₂H₂ ie. 1.5 mol H₂ per 1 mol C₂H₂

for 1 mol C₂H₂ , moles of H₂ needed can be calculated using (a) as:

moles H₂ needed = 2 mol H₂ / 1 mol C₂H₂ x 1 mol C₂H₂ 2 mol H₂

H₂ present is less than H₂ needed and hence it is limiting reactant and will be exhausted first.

(c)

molar mass of C₂H₂ = 26g/mol = 26kg/kmol

molar mass of C₂H6 = 30 g/mol = 30kg/kmol

molar mass of H₂ = 2g/mol = 2kg/kmol

converting the stoichiometric conversion in terms of mass by multiplying it with their respective molar masses

2 mol H₂ / 1 mol C₂H₂ =  2 x 2kg H₂ / 1 x 26kg C₂H₂ = 4kg H₂ / 26 kg C₂H₂

Stoichiometric conversion ratio for ethane in terms of kg mass = 2 mol H₂ / 1 mol C₂H₆ = 4kg H₂ / 30kg C₂H₆

to produce 4×10⁶ metric tons of ethane per year = 4 x 10⁹ kg

as 1 metric ton = 10³ kg

total hydrogen needed = 4kg H₂ / 30kg C₂H₆ x 4 x 10⁹ kg C₂H₆ = 5.333 x 10⁸ kg

total no. of seconds in 300 days = 300days x 24hours x 3600s = 2.592 x 10⁷ s

H₂/s = 20.57 kg/s

d)If you keep one reactant excess then it would be difficult to seperate as it will be coming in the product stream

and no one is going to buy the mixture of gases in the name of pure ethane.So to maintain product stream pure,no excess reactant is used as the remaining excess liquid will come out with the product stream and make it impure to sell.