🤓 Based on our data, we think this question is relevant for Professor Geissler's class at CAL.
Each step in the process below has a 90.0% yield.
CH4 + 4Cl2 → CCl4 + 4 HCl
CCl4 + 2 HF → CCl 2F2 + 2 HCl
The CCI4 formed in the first step is used as a reactant in the second step. If 4.50 mol of CH 4 reacts, what is the total amount of HCl (moles) produced? Assume that Cl2 and HF are present in excess.