Asked by halp
How much heat is released if 2.00 moles
CF2Cl2 reacts with 70.0 g F2 with a 75.0
percent yield?
CF2Cl2 + F2 → CF4 + Cl2
∆H for this reaction is -401 kJ/mol rxn.
CF2Cl2 reacts with 70.0 g F2 with a 75.0
percent yield?
CF2Cl2 + F2 → CF4 + Cl2
∆H for this reaction is -401 kJ/mol rxn.
Answers
Answered by
DrBob222
First we must determine the limiting reagent(LR).
mols F2 = approx 70/38 = about 1.84.
2.0 mols CF2Cl2 will produce 2 mols CF4.
1.84 mols F2 will produce 1.84 mols CF4,
Therefore, the LR is F2 and CF2Cl2 is in excess.
So the rxn releases 401 kJ for 38 g F2. You have 70 g; therefore,
401 kJ x 70/38 = ? kJ released.
mols F2 = approx 70/38 = about 1.84.
2.0 mols CF2Cl2 will produce 2 mols CF4.
1.84 mols F2 will produce 1.84 mols CF4,
Therefore, the LR is F2 and CF2Cl2 is in excess.
So the rxn releases 401 kJ for 38 g F2. You have 70 g; therefore,
401 kJ x 70/38 = ? kJ released.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.