Each step in the process below has a 70.0% 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 7.00 mol of CH4 reacts, what is the total amount of HCl produced? Assume that Cl2 and HF are present in excess.