[emilie]

Hi, I've been working on this chemistry problem for class and I'm not seeing where I'm messing up.

Here's the question:

Chlorine gas reacts with fluorine gas to form chlorine trifluoride.

Cl₂(g) + 3 F_s(g)  2 ClF₃(g)

A 2.00-L reaction vessel, initially at 298 K, contains chlorine gas at a partial pressure of 337 mmHg and fluorine gas at a partial pressure of 729 mmHg. Identify the limiting reactant and determine the theoretical yield of ClF₃ in grams.

Here's what I did:

mol Cl₂: PV = nRT

n = PV/RT = (337 mmHg x (1 atm / 760 mmHg) x 2.00 L) / (0.08206 L atm / mol K x 298K) = 0.036266 mol Cl₂

mol F₂: n = PV/RT = (729 mmHg x (1 atm / 760 mmHg) x 2.00 L) / (0.08206 L atm / mol K x 298 K) = 0.078451 mol F₂

Limiting reactant

0.036266 mol Cl₂ (3 mol F₂ / 1 mol Cl₂) = 0.109 mol F₂

Limiting reactant is F₂ since it's less than the Cl₂

Theoretical yield of ClF₃

0.078451 mol F₂ (2 mol ClF₃/3 mol F₂)(92.45 g ClF₃ / 1 mol F₂) = 4.84 g ClF₃

And this answer is not correct. What am I not seeing? Thanks!

[emilie]

Book answer is F₂ is limiting reactant and  2.84 g ClF₃ is the theoretical yield.

[Borek]

Your answer looks OK to me, must be an error in the book.