Question

A gas mixture contains 12.3 mol of Br2 and 47.1 mol of F2.

(a) Compute the mole fraction of Br2 in the mixture.

(b) The mixture is then heated, and the Br2 starts to react with the F2 to give BrF5: Br2(g) + 5F2(g) 2BrF5(g)

At a certain point in the reaction, 7.10 mol of BrF5 is present. Determine the mole fraction of Br2 in the new mixture.

Answer 1

a)

mole fraction of Br2 = mol of Br2 / Total number of moles

= 12.3 / (12.3 + 47.1)

= 0.207

Answer: 0.207

b)

mol of BrF5 present = 7.10 mol

mol of F2 reacted = (5/2)*mol of BrF5 present

= (5/2)*7.10 mol

= 17.75 mol

mol of F2 present = 47.1 - 17.75

= 29.35 mol

mol of Br2 reacted = (1/2)*mol of BrF5 present

= (1/2)*7.10 mol

= 3.55 mol

mol of Br2 present = 12.3 - 3.55

= 8.75 mol

mole fraction of Br2 = mol of Br2 present / total moles present

= 8.75 / (7.10 + 29.35 + 8.75)

= 8.75 / 45.2

= 0.194

Answer: 0.194