Finding the grams of a compound with % yield given

Question

Each step in the process below has a 80.0% yield.

The CCl4 formed in the first step is used as a reactant in the second step. If 3.50 mol of CH4 reacts, what is the total amount of HCl produced? Assume that Cl2 and HF are present in excess.

The CCl4 formed in the first step is used as a reactant in the second step. If 3.50 mol of CH4 reacts, what is the total amount of HCl produced? Assume that Cl2 and HF are present in excess.

CH4 +4Cl2 --> CCl4 + 4HCl

CCl4 + 2HF --> CCl2F2 + 2HCl

Answer 1

Since we know that Cl₂ and HF are present in excess, CH₄ is the limiting reagent. If we know 3.50 mol of CH₄ reacts then:

3.50 mol CH₄ * (4 mol HCl/1 mol CH₄) * 0.80 = 11.2 mol HCl

*Note, this is the amount of HCl produced from the first step only*

For the second step, we know that CCl₄ is still the limiting reagent, and thus, we must find out how much CCl₄ we produced from the first step:

3.50 mol CH₄ * (1 mol HCl/1 mol CH₄) * 0.80 = 2.8 mol CCl₄

Thus, 2.8 mol of CCl₄ is the limiting reagent in the second step and:

2.8 mol CCl₄ * (2 mol HCl/1 mol CCl₄) * 0.80 = 4.48 mol HCl

Summing the total amount of HCl we formed in steps 1 and 2:

11.2 mol HCl + 4.48 mol HCl = 15.68 mol HCl

We have formed a total amount of 15.68 mol HCl

Multiplying 15.68 mol HCl by its molar mass, we get:

15.68 mol * 36.46 g/mol = 571.6928 g HCl