How many grams of CIF3 form when .204 moles of F2 react with excess Cl2?
The answer must be sold with dimensional analysis. The "excess part" is confusing me! Pleas e help :/ I keep getting the wrong answers

Question

How many grams of CIF3 form when .204 moles of F2 react with excess Cl2?
The answer must be sold with dimensional analysis. The "excess part" is confusing me! Pleas e help :/ I keep getting the wrong answers

anonymous · Answer

when it says with excess Cl2, that just means that you ran out of F2 (limiting reagent) during your reaction to form ClF3 so Cl2 was not completely used up = excess of Cl2. 
But dont worry about it too much, cuz here, we are asked to find how many ClF3 formed, not Cl2.
So for this one, for every 3 mol of F2, you form 2 mol of ClF3. So the steps should look like this:

(.204mol F2) x (2mol ClF3/3mol F2) x (92.45g ClF3/ 1mol ClF3) = 37.7g ClF3

anonymous · Answer

Thank you thank you thank you! However I did this calculation, and I got 12.6g Clf3! :O Who is right? lol

anonymous · Answer

oops, a thousand apologies. you are right. 
i multiplied by 2 instead of (2/3) so it should be 12.5732g ClF3.
my baddd. haha