Physical Science Math

Question

Bismuth and fluorine react to form bismuth fluoride. If 417.96 g of bismuth reacts completely with 113.99 g of fluorine, how many grams of bismuth fluoride are formed?

Answer 1

Hello Alyssa

 

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