Question

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.

Answer 1

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.