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 utilized as a reactant in the second step. If 7.00 mol of CH4 reacts, Explain what is the total amount of HCL produced? Presume that Cl2 and HF are present in excess.