Each step in the process below has 80% 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 2.00 mol of ch4 reacts, what is the total amount of hcl produced? assume that cl2 and hf are present in the excess.
explain

8.96 mol of HCl is produced.

Explanation:

CH₄ + 4Cl₂ → CCl₄ + 4HCl   80% yield

CCl₄ + 2HF → CCl₂F₂ + 2HCl   80% yield

The first reaction:

1 mol CH₄ 4 mol HCl                  1 mol CH₄ 1 mol CCl₄

2 mol CH₄       x                          2 mol CH₄       x

x = 8 mol HCl                                        x = 2 mol HCl

The first reaction would lead to 8 mol of HCl, but as its yield is 80%, the total ammount of HCl produced is 0.8x8 = 6.4 mol HCl

The same applies for CCl₄: 0.8x2 = 1.6 mol CCl₄

The second reaction:

1 mol CCl₄ 2 mol HCl                  

1.6 mol CH₄       x                          

x = 3.2 mol HCl                                    

The second reaction would lead to 3.2 mol of HCl, but as its yield is 80%, the total ammount of HCl produced is 0.8x3.2 = 2.56 mol HCl  

Therefore, considering both reactions, 8.96 mol of HCl is produced.

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