A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 2 years ago
A chemist reacted 300 g of 1butene with excess Br2 (in CCl4) at 25 8C. He isolated 418 g of
1,2dibromobutane. What is the percent yield?
anonymous
 2 years ago
A chemist reacted 300 g of 1butene with excess Br2 (in CCl4) at 25 8C. He isolated 418 g of 1,2dibromobutane. What is the percent yield?

This Question is Closed

matt101
 2 years ago
Best ResponseYou've already chosen the best response.1Br2 is added across the double bond in 1butene to form a dihalide. The balanced reaction would be: \[C_{4}H_{8}+Br_{2} \rightarrow C_{4}H_{8}Br_{2}\] We have 300 g of 1butene, which is 300 g / 56 g/mol = 5.36 mol. According to the reaction, this means we should also get 5.36 mol of 1,2dibromobutane, which has a mass of 5.36 mol x 216 g/mol = 1158 g. The formula for percent yield is: \[percent \space yield = \frac{mass \space of \space actual \space yield}{mass \space of \space theoretical \space yield} \times 100\] So let's just plug our numbers in to find the answer: \[percent \space yield = \frac{418 \space g}{1158 \space g} \times 100 =36.1\] The percent yield for this reaction is 36.1%.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.