Question

A gas mixture contains 4.5 mol Br2 and 33.1 mol F2. (a) Compute the mole fraction of Br2 in the mixture. (b) The mixture is heated above 150°C and starts to react to give BrF5: Br2(g) 1 5 F2(g) 8n 2 BrF5(g) At a certain point in the reaction, 2.2 mol BrF5 is present. Determine the mole fraction of Br2 in the mixture at that point.

Answer 1

Answer:

a.  0.12mol

b.   0.10mol

Explanation:

Br2(g)

mole fraction of Br2(g)

=mole of Br2(g)/total number of mole of te reactants

total number of mole=4.5+33.1=37.6mol

mole fraction of Br2(g)=4.5/37.6

mole fraction of Br2(g)=0.12mol

2.  Br2(g)+5F2(g)→2BrF5(g)

if 2.2mol of BrF5 is produced , goin by te equation above

Br2=1.1mol

5F2=5.5mol

after the reaction, the following will be present

br2, 4.5-1.1=3.4mol

F2, 33.1-5.5=27.6mol

BrF5=2.2mol

mol fraction of Br2=3.4/(3.4+27.6+2.2)

mol fraction of Br2=3.4/33.2

0.10mol