Question

Chloride reacts with propylene to form allyl chloride (C3H5Cl) and dicholopropane (C3H6Cl2) according to the following reactions: The product gases contain 141 mol Cl2, 651 mol C3H6, 4.6 mol C3H5Cl, 24.5 mol C3H6Cl2 and 4.6 mol HCl. a) How much chloride and propylene were fed to the reactor? b) What is the limiting reactant? c) What is the excess percentage? d) What is the fractional conversion of Cl2? e) What was the selectivity of C3H5Cl relative to C3H6Cl2? f) What was the ratio of C3H5Cl produced expressed in grams to the grams of C3H6 fed to the reactor?

Answer 1

The balanced reactions are

C3H6 (g) + Cl2 (g) = C3H5Cl (g) + HCl (g)

C3H6 (g) + Cl2 (g) = C3H6Cl2 (g)

In product gas

Moles of Cl2 = 141 mol

Moles of C3H6 = 651 mol

mol of C3H5Cl = 4.6 mol

mol of C3H6Cl2 = 24.5 mol

mol of HCl = 4.6 mol

From the first reaction

Moles of C3H6 consumed = moles of Cl2 consumed = moles of C3H5Cl produced = mol of HCl produced

Moles of C3H6 consumed = moles of Cl2 consumed = 4.6 mol

From the second reaction

Moles of C3H6 consumed = moles of Cl2 consumed = moles of C3H6Cl2 produced

Moles of C3H6 consumed = moles of Cl2 consumed = 24.5 mol

Total Moles of C3H6 consumed = 4.6 + 24.5 = 29.1 mol

Total Moles of Cl2 consumed = 4.6 + 24.5 = 29.1 mol

Part a

Moles of C3H6 fed into the reactor = moles of C3H6 consumed + moles of C3H6 in product gas

= 29.1 + 651 = 680.1 mol

Moles of Cl2 fed into the reactor = moles of Cl2 consumed + moles of Cl2 in product gas

= 29.1 + 141 = 170.1 mol

Part b

Stoichiometric Molar ratio of C3H6 : Cl2 = 1 : 1

Actual molar ratio of C3H6 : Cl2 = 680.1 : 170.1 = 4 : 1

Moles of C3H6 fed > moles of Cl2 fed

Limiting reactant = Cl2

Excess reactant = C3H6

Part c

% C3H6 in excess

= (Moles of C3H6 fed - moles of Cl2 fed) x 100 / (moles of Cl2 fed)

= (680.1 - 170.1) x 100 / (170.1)

= 299.82 %

Part d

Fractional conversion of Cl2 = moles reacted / moles fed

= 29.1 / 170.1

= 0.171