Chorine and fluorine react to form gaseous chlorine trifluoride. In an experiment, you started with 1.75 moles of chlorine and 8.68 moles of fluorine.

Question

___Cl2 + ____F2 ---> ___ClF2

a. What is the limiting reactant?

b. What mass of chlorine triflouride can be made?

c. Which reactant is in excess?

d. What mass of the excess reactant remains unused?

Answer 1

Cl2 + 3F2 ==> 2ClF3 (note: chlorine trifluoride is ClF3, not ClF2)

For Cl2: 1.75/1 = 1.75

For F2: 8.68/3 = 2.89

Cl2 is limiting

mass ClF3 = 1.75 mol Cl2 x 2 mol ClF3/mol Cl2 x mokar mass ClF3 = answer in grams

Since Cl2 is limiting, F2 is in xs

moles F2 present - 3x1.75 = moles F2 remaining

moles F2 remaining x mokar mass F2 = mass F2 remaining