Question

Each step in the following process has a yield of 90.0 %. CH 4 + 4 Cl 2 ⟶ CCl 4 + 4 HCl CCl 4 + 2 HF ⟶ CCl 2 F 2 + 2 HCl The CCl 4 formed in the first step is used as a reactant in the second step. If 3.00 mol CH 4 reacts, what is the total amount of HCl produced? Assume that Cl 2 and HF are present in excess.

Answer 1

When 3.00 mol of CH4 reacts in a two-step process with each step having a 90% yield, a total of 16.86 mol of HCl is produced. This includes 12.00 mol from the first step and 4.86 mol from the second step.

Step-by-step explanation:

To calculate the total amount of HCl produced when 3.00 mol CH4 reacts, we need to consider the yields of both steps in the process:

1. CH4 + 4 Cl2 → CCl4 + 4 HCl
2. CCl4 + 2 HF → CCl2F2 + 2 HCl

With a 90% yield for each step, we first calculate the moles of CCl4 formed from CH4:
3.00 mol CH4 × 90% = 2.70 mol CCl4

Now we calculate the moles of HCl from the first step (4 moles of HCl for every 1 mole of CH4):
3.00 mol CH4 × 4 mol HCl/mol CH4 = 12.00 mol HCl

For the second step, we calculate the moles of HCl produced (2 moles of HCl for every 1 mole of CCl4) considering the 90% yield:
2.70 mol CCl4 × 2 mol HCl/mol CCl4 × 90% = 4.86 mol HCl

The total HCl produced is the sum of HCl from both steps:
12.00 mol HCl + 4.86 mol HCl = 16.86 mol HCl