Question

Calculate:
In the following equation, 417.96 g of Bi (bismuth) reacts completely with 200 g of F (fluorine.) How many grams of BiF (little three below F) are formed?
2 Bi + 3 F (little 2 under F) ---> 2 BiF (little three below F)

Answer 1

This is a limiting reagent (LR) problem and I know that because amounts are given for BOTH reactants. I solve these the long way--there is a shorter way but I think it's harder to explain, especially by typing.
2Bi + 3F2 ==> 2BiF3

mols Bi = grams/molar mass = ?
mols F2 = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols Bi to mols BiF3.
Do the same to convert mols F2 to mols BiF3.
It is quite likely that the two values will not agree; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that smaller value is the LR.
Take the smaller value and convert that to grams. g = mols x molar mass. = ?grams BiF3

Answer 2

500