8Fe+S8→8FeS If we begin with 293 g Fe and 17.2 g S8, how many grams of FeS will we create?

Answer 1

8Fe + S₈ ==> 8FeS

Whenever you are given the amounts of BOTH reactants, you must find which, if either, is limiting.

For Fe we have:

293 g Fe x 1 mol Fe/55.85 g x 8 mol FeS/8 mol Fe x 87.92 g FeS/mol FeS = 461 g FeS formed

For S₈ we have:

17.2 g S₈ x 1 mol S₈/256.5 g x 8 mol FeS/1 mol S₈ x 87.92 g FeS/mol FeS = 47.2 g FeS formed

Clearly, S₈ is in limiting supply and the maximum mass of FeS will be 47.2 g