Each step in the process below has a 70.0% yield

Reference no: EM13847712

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.

Reference no: EM13847712

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