2Na3P + 3CaF2 ===> 6NaF + Ca3P2 Balanced Equation
moles of Na3P = 107.00 g x 1 mole/99.94 g = 1.071 moles
Assuming CaF2 is in excess, as the problem does not mention how much is present, then 1.071 moles Na3P will be limiting.
1.071 moles Na3P x 6 moles NaF/2 moles Na3P = 3.213 moles NaF will be produced
Mass of NaF produced = 3.213 moles x 41.99 g/mole = 134.9 grams NaF produced (to 4 significant figures)