Bromine (br2) is produced by reacting hbr with o2, with water as a byproduct. the o2 is part of an air (21 mol % o2, 79 mol % n2) feed stream that is flowing sufficiently fast to provide 25% excess oxygen ("excess" has a precise meaning in process analysis: in this case there is 25% more oxygen than the amount needed to completely react with the limiting reactant). the fractional conversion of hbr is 78%.
a) show the degree of freedom analysis. be as specific as possible about labeling the unknowns and completely write out all of the independent equations.
b) calculate the composition (mole fractions) of the product stream.

The mole fractions:

Explanation:

The reaction described is:

The limiting reactant is the HBr (oxygen is in excess).

a) The mass (in moles) balance for this sistem:

(the 0.78 is because of the fractional conversion)

(the 1.25 is because of the oxygen excess)

There is only one degree of freedom in this sistem, you can either deffine the moles of HBr you have or the moles of Br2 you want to produce. The other variables are all linked by the equations above.

b) Base of calculation 100 mol of HBr:

The mole fractions:

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )