Manganese trifluoride, MnF3,can be prepared by the following reaction.

Question

2 MnI2 (s) + 13 F2 (g) ---> 2 MnF3(s) + 4 IF5(l)
What is the minimum number of grams of F2 that must be used to react with 12.0 g of MnI2 if the overall yield of MnF3 is no more than 75% of theory?

Answer 1

12.0g MnI2/0.75 = ? g MnI2 to start accounting for the 75% yield = approx 16

mols MnI2 = grams/molar mass = 16/about 300 = ?
Convert to mols F2. That's
? mols MnI2 x (13mols F2/2 mols MnI2) = ? x 13/2 = x mols MnI2.
Convert to grams. g = mols x molar mass