You want to first write the balanced chemical reaction between Bismuth and Fluorine to obtain the mole ratios between the 2 reactants and the product, bismuth fluoride.
2 Bi + 3 F2 → 2 BiF3
Molar mass of Bi = 208.98 g/mol
Molar mass of F2 = 38.00 g/mol
Molar mass of BiF3 = 265.98 g/mol
Since we have 2 reactants, we want to determine whether Bismuth or Fluorine is the limiting reagent. For this, we need to convert grams of each reactant to moles of product.
Bi = 419.96 g Bi x 1 mol Bi / 208.98 g Bi x 2 mol BiF3 / 2 mol Bi = 2.0096 mol BiF3 (to 5 significant figures based on mass of Bi)
F2 = (113.99 g F2 x 1 mol F2 / 38.00 g F2 x 2 mol BiF3 / 3 mol F2 = 1.9998 mol BiF3
As you can see that Fluorine is the limiting reagent, therefore, the amount of BiF3 that forms will be based on the starting mass of F2.
Finally, take the moles of BiF3 formed from F2 and convert it grams of BiF3 using its molar mass.
1.9998 mol BiF3 x 265.98 g BiF3 / 1 mol BiF3 = 531.91 g BiF3