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.

Cl

_{2}(g) + 3 F

_{s}(g)

2 ClF

_{3}(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 ClF

_{3} in grams.

Here's what I did:

**mol Cl**_{2}: PV = nRT

n = PV/RT = (337

~~mmHg~~ x (1

~~atm~~ / 760

~~mmHg~~) x 2.00

~~L~~) / (0.08206

~~L~~ ~~atm~~ / mol

~~K~~ x 298

~~K~~) = 0.036

266 mol Cl

_{2}**mol F**_{2}: n = PV/RT = (729

~~mmHg~~ x (1

~~atm~~ / 760

~~mmHg~~) x 2.00

~~L~~) / (0.08206

~~L~~ ~~atm~~ / mol

~~K~~ x 298

~~K~~) = 0.078

451 mol F

_{2}Limiting reactant0.036

266 mol Cl

_{2} (3 mol F

_{2} / 1 mol Cl

_{2}) = 0.109 mol F

_{2}Limiting reactant is F

_{2} since it's less than the Cl

_{2}Theoretical yield of ClF_{3}0.078

451

~~mol F~~_{2} (2

~~mol ClF~~_{3}/3

~~mol F~~_{2})(92.45 g ClF

_{3} / 1

~~mol F~~_{2}) = 4.84 g ClF

_{3}And this answer is not correct. What am I not seeing? Thanks!