A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
150.0 g of AsF3 was reacted with 180.0 g of CCl4 to produce AsCl3 and CCl2F2. The theoretical yield of CCl2F2 produced, in moles, is?
a]0.7802 mol
b]1.274 mol
c]1.170 mol
d]0.5685 mol
e]1.705 mol
anonymous
 5 years ago
150.0 g of AsF3 was reacted with 180.0 g of CCl4 to produce AsCl3 and CCl2F2. The theoretical yield of CCl2F2 produced, in moles, is? a]0.7802 mol b]1.274 mol c]1.170 mol d]0.5685 mol e]1.705 mol

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0AsF3 + CCl4 = AsCl3 + CCl2F2 you need all the mol number of each substance of both sides to be equal. start with F 3F on the left, 2F on the right so time 2 to 3F on the left, time 3 to 2F com the right, we get 2AsF3 + CCl4 = AsCl3 + 3CCl2F2 Now, there are 2 As on the left, 1 As on the right so time 2 to As on the right to balance 2AsF3 + CCl4 = 2AsCl3 + 3CCl2F2 now there are 4Cl left, 12Cl right, also 1C left, 3C right so time 3 to CCl4 2AsF3 + 3CCl4 = 2AsCl3 + 3CCl2F2 that's the balance part

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when u get used to it, u will just put number in there by trying, shouldn't take more than 30 secs.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wow... what do I do afterwards?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now group what information you have M: atomic mass, u know this. n: mol number 150.0 g of AsF3 MAs=75g/mol MF=19g/mol MAsF3=75+19*3=132g/mol nAsF3=150/132=1.136mol 180.0 g of CCl4 MC=12 MCl=35.5 MCCl4=12+35.5*4=154 nCCl4=180/154=1.169mol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you so much :) Now you can go back to bed :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02AsF3 + 3CCl4 = 2AsCl3 + 3CCl2F2 theoretically 2mol 3mol don't need 3mol reality 1.136 1.169 should be 1.132 > 1.704 try with CCl4 1.169 blabla, so you can take over from here right? i'll leave the rest for you
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.