Question

Carbon tetrachloride, CCl₄, is a solvent that was once used in large quantities in dry cleaning. Because it is a dense liquid that does not burn, it was also used in fire extinguishers. Unfortunately, its use was discontinued because it was found to be a carcinogen. It was manufactured by the following reaction:
$CS_2+3Cl_2 \longrightarrow CCl_4+S_2Cl_2$
The reaction was economical because the byproduct disulfur dichloride, S₂Cl₂, could be used by industry in the manufacture of rubber products and other materials.
a. What is the percentage yield of CCl₄i if 719 kg is produced from the reaction of 410. kg of CS₂?
b. If 67.5 g of Cl₂ are used in the reaction and 39.5 g of S₂Cl₂ is produced, what is the percentage yield?
c. If the percentage yield of the industrial process is 83.3%, how many kilograms of CS₂ should be reacted to obtain 5.00 × 10⁴ kg of CCl₄? How many kilograms of S₂Cl₂ will be produced, assuming the same yield for that product?

Answer 1

Answer:

The answers to the question are

a. Percentage yield  = 86.804 %

b. Percentage yield =92.2 %

c. 3.09×10⁶ kg of CS₂ should be reacted to obtain 5.00 × 10⁴ kg of CCl₄ at 83.3 % yield

c. (i) 4.39×10⁶ kg of S₂Cl₂ will be produced alongside at 83.3 %yield

Explanation:

a.To solve the question, we note that in the chemical reaction

1 mole of CS₂ reacts with 3 moles of Cl₂ to produce 1 mole of CCl₄ and 1 mole of S₂Cl₂

Mass of CCl₄ produced = 719 kg (Actual yield)

Mass of CS₂ in the reaction = 410 kg

Molar mass of CCl₄ = 153.82 g/mol,

Number of moles of CCl₄ =$\frac{Mass}{Molar mass}$ = 719/153.82 = 4.674 moles

Molar mass of CS₂ = 76.139 g/mol

Number of moles of CS₂ =$\frac{Mass}{Molar mass}$ = 410/76.139 = 5.385 moles

Therefore theoretical yield of CCl₄= 5.385 moles×153.82 g/mol

= 828.303 kg

Percentage yield = $\frac{Actual Yield}{Theoretical Yield} *100$ → $\frac{719}{828.303} *100$ = 86.804 %

b. Mass of the Cl₂ used in the reaction = 67.5 g

Mass of S₂Cl₂ produced =39.5 g (Actual yield)

Molar mass of Cl₂ = 70.906 g/mol

Number of moles of Cl₂ =67.5/70.906 = 0.952 moles

Molar mass of S₂Cl₂ = 135.04 g/mol

Number of moles of S₂Cl₂ =39.5/135.04 = 0.293 moles

However here  3 moles of  Cl₂ produces 1 mole of S₂Cl₂, therefore 0.952 moles of Cl₂ will produce 0.952/3 or 0.317 moles of S₂Cl₂

Therefore actual yield = 0.317 moles ×135.04 g/mol = 42.851 g

Percentage yield = 39.5/42.851×100 =92.2 %

c. To produce 5.00 × 10⁴ kg at 83.3 % yield

We have Percentage yield =   $\frac{Actual Yield}{Theoretical Yield} *100$, we have

Theoretical yield =  $\frac{Actual Yield}{Percentage Yield} *100$ = $\frac{500*10^4}{83.3} *100$ = 600.24 × 10⁴ kg

Number of moles of CCl₄ in 600.24 × 10⁴ kg of CCl₄ is given as

600.24 × 10⁷ g/153.82 g/mol, = 39022240.023 moles

Since 1 mole of CS₂ is required to produce 1 mole of CCl₄. then 39022240.023 moles of CS₂ is required to produce 39022240.023 moles of CCl₄ theoretically

mass of required CS₂ = Number of moles ×Molar mass of CS₂ = 39022240.023 moles  × 76.139 g/mol = 3088181053.204 = 3.09×10⁶ kg

Therefore 3.09×10⁶ kg should be reacted to obtain 5.00 × 10⁴ kg of CCl₄

Mass of 39022240.023 moles of  S₂Cl₂ is given by

39022240.023 moles × 135.04 g/mol = 5269563292.7466 g  which at 83.3 % yield gives 0.833 ×5269563292.7466 g = 4389546222.8579 g

= 4389546.2228579 kg = 4.39×10⁶ kg of S₂Cl₂