Question

The chlorination of methane occurs by the following reaction:

CH4 + Cl2 → CH3Cl + HCl

The conversion of the limiting reactant is 67%, and the feed composition in mole percent is 40% CH4, 50% Cl2, and 10% N2.

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

(b) Draw a fully labeled process flowchart, and perform a DOF analysis using all 3 methods (molecular species method, atomic species method, and extent of reaction method).

(c) Using any method you wish, solve for all your unknowns.

Answer 1

The chlorination of methane occurs by the following reaction:

CH4 + Cl2 → CH3Cl + HCl

The conversion of the limiting reactant is 67%, and the feed composition in mole percent is 40% CH4, 50% Cl2, and 10% N2.

Basis : 100 mole of feed containing 40 mole of CH4, 50 moles of Cl2 and 10 moles of N2.

1 mole of CH4 as per the reaction requires 1 mole of Cl2 to produce 1 mole of CH3Cl.

Molar ratio of reactants ; 1:1 Supplied ratio 40 :50 =1:1.25. So Cl2 is excess and CH4 is limiting. Percentage excess= 100*(50-40)/40= 25%

Unknowns : Mole of CH3Cl, Moles of HCl.

The conversion of limiting reactants is 67%. Hence CH4 converted= 40*0.67=26.8 moles

Cl2 converted= 26.8 moles, CH4 remaining = 40-26.8= 13.2 moles Cl2 remaining= 50-26.8= 23.2 moles

Number of unknown in the products : CH3Cl, HCl, N2, CH4, Cl2

CH4 Reacted and Cl2 reacted are related by the extent of reaction of limiting reactants

Extent of limiting reactant converted= mole of CH4 as well as mole of Cl2 reacted. This equation will help in also calculating the mole of CH3Cl and HCl formed. Remaining CH4 = Moles taken initially – moles of CH4 consumed, Moles of Cl2 = moles of Cl2 taken- mole of Cl2 consumed.

Mole of N2 in feed =mole of nitrogen in in the products.