when bismuth reacts with fluorine, bismuth (III) fluoride is produced. if 5.00g of bismuth reacts with excess flourine to produce 4.00g of BiF3, what is the theoretical yield?

Answer 1

Bi + F ---> BiF₃

Balance the equation:

Bi + 3F ---> BiF₃

Note that theoretical yield just means how much you should be able to get according to the math of the equation, nothing to do with the experimental value.

Molar Mass (MM) of bismuth (Bi) = 209 grams/mole (from periodic table)

Excess fluorine means we can assume we have an infinite amount of it. The number doesn't matter.

Molar Mass (MM) of bismuth (III) Fluoride (BiF₃) = MM of bismuth + MM of fluorine (F)*3 = 209 + 38 + 38 + 38 = 323 grams/mole

5.00 grams of bismuth (Bi) is how much in moles? Divide grams by Molar Mass (MM)

5.00/209 = 0.024 moles

Even if you change the number of moles in an equation, the mole ratios stay the same!

If we start with 1 mole of bismuth (Bi) and end with 1 mole of bismuth (III) fluoride (BiF₃), that's a 1:1 ratio

A 1:1 ratio for 0.024 moles means bismuth (III) fluoride will also have 0.024 moles.

1:1 is the same as 0.024:0.024

0.024 moles of bismuth (III) fluoride (BiF₃) is how many grams? Multiply Molar Mass (MM) by Moles

323*0.024 = 7.75 grams

This is the theoretical yield!