How many grams of CIF3 form form from 130.0g of Cl2 when F2 is in excess?
I did this: 130.0g Cl2 x (1mol Cl2/70.9 g Cl2) (2mol CIF3/1mol Cl2) (195 g CIF3/ 1mol CIF3).

I came up with 715g CIF3, but that is way to big! The answer is supposed to be 339 g CIF3. Please tell me what I need to do to fix this? I need to answer using dimensional analysis.

Question

How many grams of CIF3 form form from 130.0g of Cl2 when F2 is in excess?
I did this: 130.0g Cl2 x (1mol Cl2/70.9 g Cl2) (2mol CIF3/1mol Cl2) (195 g CIF3/ 1mol CIF3). 

I came up with 715g CIF3, but that is way to big! The answer is supposed to be 339 g CIF3. Please tell me what I need to do to fix this? I need to answer using dimensional analysis.

anonymous · Answer

You did everything right step-wise, you just messed up on the molar mass for ClF3. Here's what i got:
Cl = 35.45g
3F = 57g
Add them up and you'll have have 92.45g for every mol of ClF3. So just change the last part of your work and you should get:

(130g Cl2)(1mol Cl2/70.9 g Cl2)(2mol ClF3/1mol Cl2)(92.45g ClF3/1mol ClF3) = 339gClF3