ques 22 KM Each step in the process below has a 700 yield CH

ques 22 K-M

Each step in the process below has a 70.0% yield.

CH4+4CI2----> CCI4+4HCI

CCI4+2HF----->CCI2F2+2HCI

The CCl4 formed in the first step is used as a reactant in the second step. If 9.00 mol of CH4 reacts, what is the total amount of HCl produced? Assume that Cl2 and HF are present in excess.

Solution

(9.00 mol CH4) x (1/1) x (0.70) = 6.30 mol CCl4
(9.00 mol CH4) x (4/1) x (0.70) = 25.2 mol HCl

CCl4 + 2 HF CCl2F2 + 2 HCl

(6.30 mol CCl4) x (2/1) x (0.70) = 8.82 mol HCl

25.2 mol + 8.82 mol = 34.02 mol HCl from both steps