Question

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

Answer 1

Answer:

Formula weight of fluorine is 19. Formula weight of Bromine is 209 So formula weight of BiF3 is 209+ 3(19)= 266 g/mole. Given Bi of 418g, so 2 moles of Bi. F2 is 200g , so 5.25 moles of F2. Hence 266g/mole *(2 moles)= 532 g of BiF3 is formed

Answer 2

The weight of formed Bismuth fluoride is 531.94g.

Bismuth and Fluorine react to form Bismuth fluoride. If 417.96g of Bismuth react completely with 113.99 g of Fluorine.

We have to find the weight of formed Bismuth fluoride.

solution :

here,

• mass of Bismuth = 417.96 g
• mass of fluorine = 113.99 g

Atomic mass of Bismuth = 208.98 g/mol

∴ no of moles of Bismuth = 417.96/208.98 = 2

similarly,

atomic mass of Fluorine = 18.99 g

∴ no of moles of Fluorine = 113.99/18.99 = 6

Bismuth and Fluorine reacts as shown below ;

2Bi + 3F₂ ⇒ 2BiF₃

It is clear that 2 moles of Bismuth reacts with 3 moles of Fluorine and form 2 moles of Bismuth fluoride.

here you see, we have only 2 moles of Bismuth instead of 4 moles for 6 moles of Fluorine (in excess amount).

so, Bismuth is limiting reagent.

now we should give priority to Bismuth.

2 moles of Bismuth form 2 moles of BiF₃.

∴ mass of Bismuth fluoride = 2 × molecular weight of Bismuth fluoride

= 2 × 265.97 g

= 531.94 g

Therefore the weight of formed Bismuth fluoride is 531.94g.