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

Answer:

4.86 moles of HCl

Explanation:

1. First write the balanced chemical equations involved in the process:

$CH_{4}+_4Cl_{2}=CCl_{4}+_4HCl$

$CCl_{4}+_2HF=CCl_{2}+F_{2}+_2HCl$

2. Calculate what amount of $CCl_{4}$ is formed in the first reaction.

$3.00molesCH_{4}*\frac{1molCCl_{4}}{1molCH_{4}}=3.00molesCCl_{4}$

As the yield of each reaction is 90.0%, the amount of $CCl_{4}$ produced is the following:

$3.00molesCCl_{4}*0.90=2.7molesCCl_{4}$

3. Calculate the amount of HCl produced.

$2.7molesCCl_{4}*\frac{2molHCl}{1molCCl_{4}}=5.4molesHCl$

The total amount of HCl produced with a 90.0% yield is:

$5.4molesHCl*0.90=4.86molesHCl$