A reactor is supplied with 2 streams both at the rate 5 mole, one has 40% Cl2 and 60% N2, and another has pure C2H6, if the product has C2H5Cl, HCl and N2 at the rate 20 mole, what is the percentage of HCl in products?

Correct Answer: 15%
The reaction is C2H6 + Cl2 -> C2H5Cl + HCl, Let the fraction of C2H5Cl, HCl and N2 be x, y and (1 – x – y) respectively. Element balances, Cl: 0.4*5(2) = x*10(1) + y*10(1), => x + y = 0.4, C: 5(2) = x*20(2), => x = 0.25, => y = 0.4 – 0.25 = 0.15, => percentage of HCl = 15%.