Question

Each step in the following process has a yield of 70.0%. 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 2.00 mol CH4 reacts, what is the total amount of HCl produced? Assume that Cl2 and HF are present in excess. moles HCl : mol

Answer 1

Answer:

n(HCl)=1.96 mol

Explanation:

CH4+4Cl2⟶CCl4+4HCl

CCl4+2HF⟶CCl2F2+2HCl

With ideal yields we will end up with 4 moles of HCl.

With 70% yields on every stage

n(HCl)=0.7*0.7*4=1.96 mol