Agas 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.

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

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

increase masses, decrease separation