If 107.00 g of Na3P react how many grams of NaF would be produced? (Show Work Please)

Question

If 107.00 g of Na3P react how many grams of NaF would be produced? Formula 2Na3P + 3CaF2 = 6NaF + Ca3P2 (already balanced)

Answer 1

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)