Question

The solvent dichloromethane, CH2Cl2, can be prepared by reaction of methane with chlorine gas, according to the following equation. This reaction was performed using 5.00 g of methane and 15.0 g of chlorine gas.

CH4 + Cl2 —> CH2Cl2 + HCl

a. What is the limiting reactant? Show all work.

b. How many grams of each product are formed?

c. If 15.6 g of dichloromethane are formed, what is the percent yield of the reaction? Is this acceptable? Why or why not?

Answer 1

balance

CH4 + 2Cl2 —> CH2Cl2 + 2HCl

a)

mol of methane = mass/MW = 5/16 = 0.3125

mol of Cl2 = mass/MW = 15/70.9 = 0.2115

ratio is 1:1 so

0.2115 mol of Cl2 --> 0.2115 /2 = 0.10575 mol of CH4 required

then Cl2 limits reaction

b)

mol of CH4 reacted --< 0.10575 mol

mol of CH2Cl2 --> 0.10575

mol of HCl = 0.10575*2 = 0.2115

mass of CH2Cl2 = mol*MW = 0.10575*84.9326 = 8.9816 g

mass of HCl = mol*MW = 0.2115*36.5= 7.719 g

c)

mass of CH2Cl = 15.6

% yield = rea / theoretical * 100% = 15.6/8.9816 *100 = 173.68 %

clearly, not possible, we cna't create mass

original balance is 5 g + 15 g = 20 g, you cant simply form 15.6 g of product alone