Answered

Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

4. What is the percent yield of a reaction that produces 12.5 g CF2Cl2 from 32.9 g of CCl4 and excess HF

Sagot :

anfabba15

Answer:

Percent yield = 48.3%

Explanation:

The reaction is:

CCl₄  +  2HF → CF₂Cl₂ + 2HCl

1 mol of CCl₄ reacts with 2 moles of hydrofluoric acid in order to produce 1 mol of CF₂Cl₂ and 2 moles of hydrogen chloride.

HF is in excess, so the limiting reagent is the CCl₄.

We convert mass to moles:

32.9 g . 1mol / 153.8g = 0.214 moles

Ratio is 1:1. In conclussion: 0.0813 moles of CCl₄ can produce 0.0813 moles of CF₂Cl₂. We convert moles to mass, to determine the theoretical yield:

0.214 mol . 120.91g /mol = 25.8 g

Percent yield = (Yield produced /Theoretical yield) . 100

Percent yield = (12.5 g/ 25.8g) . 100 = 48.3%

Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.