Question

Chlorine gas reacts with fluorine gas to form chlorine trifluoride. Cl2 (g) + 3 F2 (g) → 2 ClF3 (g) A 1.65 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 741 mmHg . What is the pressure of ClF3 in the reaction vessel after the reaction? Enter your answer numerically, in terms of mmHg.

Answer 1

pressure of Cl2 = 337 mmHg = 0.443 atm

volume = 1.65 L

temperature = 298 K

moles of Cl2 gas n = P V / R T

                              = 0.4434 x 1.65 / 0.0821 x 298

                              = 0.0299 mol

pressure of F2 = 741 mmHg = 0.975 atm

moles of F2 gas = 0.975 x 1.65 / 0.0821 x 298

                          = 0.06576

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

   1           3                                      2

0.0299     0.0658

here limiting reagent is F2.

3 mol F2 -----------------> 2 mol ClF3

0.0658 mol F2   --------------> ??

moles of ClF3 = 0.0438 mol

pressure of ClF3 = 0.0438 x 0.0821 x 298 / 1.65

                            = 0.65 atm

pressure of ClF3 = 494 mmHg