Question

Chlorine gas reacts with fluorine gas to form chlorine trifluoride.

Cl2(g) + 3 F2(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 deter- mine the theoretical yield of ClF3 in grams.

I got 4.84g ClF3 for this problem, but the textbook answer is 2.84g. Can anyone tell me where I went wrong, or show me how to get that answer?

Answer 1

                 Cl2 (g)         +     3 F2 (g) -----> 2 ClF3 (g)

Initially   337 mm Hg 729 mm Hg          0

Volume of vessel = 2 L

number of moles of Cl2 n = PV/RT

                       n = (337/760)*2 / (0.0821 * 298)

                       n = 0.03625 moles

number of moles of F2 n = PV/RT

                      n = (729/760)*2 / (0.0821 * 298)

                      n = 0.07841 moles

1 mole of Cl2 requires 3 moles of F2

0.03625 moles of Cl2 requires 3 * 0.03625 moles of F2

                             = 0.1087 moles

but we have only 0.07841 moles hence F2 is limiting reagent.

from the equation 3 moles of F2 produces 2 moles of ClF3

0.07841 moles of F2 produces 0.07841 * 2/3 moles of ClF3

                            = 0.05227 moles of ClF3

mass of ClF3 = 0.05227 * 92.45 g/mol

mass of ClF3 = 4.8323 g