Question

a chemist carries out the reaction of 0.250 mol of f2 with excess of cl2. At 250 degress C and 1.00 atm, what is the maximum volume of CLF3 in liters that the reaction can produce?

Answer 1

3F2 + 3Cl2 ==> 2ClF3

mols F2 = 0.250
mols ClF3 = 0.250 x (2 mols ClF3/3 mols F2) = 0.167 mols F2.
Convert this to volume using PV = nRT

Answer 2

PV=nRT

(1)(x)= (.250)(0.0821)(523)

1x= 10.73

Answer 3

Actually, the answer is 7.16 L