Stoichiometry mass of feed required?

Question

Ethene and chlorine react to form dichloroethane: C2H4 + Cl2 ----> C2H4Cl2, But there is a side reaction where some of chlorine and ethene react: C2H4 + 2Cl2----->C2H3Cl3 + HCl. The feed reactor contains 1.15kmol of C2H4 per kmol of Cl2. The fractional conversion of Cl2 is 100% and the selectivity of ...

Answer 1

This seems to be a ChemE type question, but as an organic chemist, I'll give it a shot.

The way I wiuld look at this is to consider the two reactions together.

First reaction, the "normal one":

C2H4 + Cl2 ----> C2H4Cl2

Second reaction, the side reaction:

C2H4 + 2Cl2----->C2H3Cl3 + HCl.

We are given that the C2H4Cl2/C2H3Cl3 ratio is 18. We can use that as a multiplication factor to add the two reactions together:

18*(C2H4 + Cl2 → C2H4Cl2)
1*(C2H4 + 2Cl2 → C2H3Cl3 + HCl)
--------------------------------------...
19C2H4 + 20Cl2 → 18C2H4Cl2 + C2H3Cl3 + HCl

Now, we can use this balanced equation that includes the side product to calculate the amounts of reactants.

Some needed conversions and quantities:
1 tonne = 1000 kg
100,000 tonnes = 100,000,000 kg = 1.0*10^8 kg
MW(C2H4) = 28.05 g/mol = 28.05 kg/kmol
MW(Cl2) = 70.906 g/mol = 70.906 kg/kmol
Mw(C2H4Cl2) = 98.96 g/mol = 98.96 kg/kmol

Calculate the moles of product (100,000 tonnes of C2H4Cl2)

(1.0*10^8 kg of C2H4Cl2)/98.96 kg/kmol = 1010509 kmol of C2H4Cl2
(1.010509*10^6 kmol of C2H4Cl2)

The amount of chlorine is the stoichiometric amount:

(1010509 kmol of C2H4Cl2)*((20 mol Cl2)/(18 mol C2H4Cl2)) = 1122788 kmol of Cl2

Mass of Cl2 = (1122788 kmol of Cl2)*(70.906 kg/kmol) = 79612413 kg of Cl2
(7.9612413*10^7 kg of Cl2)
(79612413 kg of Cl2)*(1 tonne/1000 kg) = 79,612 tonnes of Cl2

In calculating the mass of ethene, we need to take in consideration the (1.15 mol C2H4)/(1 mol Cl2) feed ratio.
Stoichiometric ratio: 19C2H4 + 20Cl2 → 18C2H4Cl2 + C2H3Cl3 + HCl
20*1.15 = 23, so
Reactor feed ratio: 23C2H4 + 20Cl2 → 18C2H4Cl2 + C2H3Cl3 + HCl

Now we can calculate the moles of ethene needed.

(1010509 kmol of C2H4Cl2)*((23 mol C2H4)/(18 mol C2H4Cl2)) = 1291206 kmol of C2H4
(1.291206*10^6 kmol of C2H4)

Mass of C2H4 = (1291206 kmol of C2H4)*(28.05 kg/kmoll) = 36218337 kg of C2H4
(3.6218337*10^7 kg of C2H4)
(36218337 kg of C2H4)*(1 tonne/1000 kg) = 36,218 tonnes of C2H4