Question

Each step in the process below has a 60.0% yield CH4 + 4Cl2 > CCl4 + 4HCl CCl4 + 2HF > CCl2F2 + 2HCl The CCl4 formed in the first step is used as a reactant in the second step. If 3.50 mol of CH4 reacts, what is the total amount of HCl produced? Assume that Cl2 and HF are present in excess.

Answer 1

1) CH4 + 4 Cl2 ---> CCl4 + 4 HCl

2) CCl4 + 2 HF ----> CCl2F2 + 2 HCl

Given,

3.5 mol CH4

Step1: moles of HCL and CCl4 produced

From(1)

1 mol of CH4 can produce 4 mol HCl

3.5 mol CH4 can produce 4x3.5= 14 mol HCl

Thus actual moles of HCl produced= 14 x 60/100. ( 60 % yield)

=8.4 mol HCl

1 mol of CH4 can produce 1 mol CCl4

3.5 mol of CH4 can produce 3.5 mol of CCl4

Actual mol of CCl4 = 3.5 x 60/100 = 2.1 mol

From(2)

1 mol CCl4 can produce 2 mol of HCl

2.1 mol CCl4 can produce 2 x 2.1 = 4.2 mol HCl

Actual mol of HCl= 4.2 x 60/100 = 2.52

Step2: total mol of HCl produced

Total mol= 8.4 + 2.52 = 10.92 moles of HCl