Answer:
Thanks for the points
Explanation:
the answer is 3
hope this works
A beaker
413.20 ppm
n(HCl)=1.96 mol
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