In an experiment to determine how to make sulfur trioxide, a chemist combines 32.0 g of sulfur with 50.0 g of oxygen. she finds that she has made 80.0 g of sulfur trioxide and has 2.0 g of oxygen leftover in the end. how would this chemist make 100.0 g of sulfur trioxide so that she has no leftover sulfur or oxygen?

The chemist needs to react 40 g of sulfur with 60 g of oxygen to make 100 g of sulfur trioxide.

Explanation:

2S (s) + 3O₂ (g) → 2SO₃ (g)

64g    + 96g     →  160 g

32g    + 48g     →   80 g

   x     +     y      →  100 g

1 mol SO₃ ___ 80g

     n 100g

         n = 1.25 mol SO₃

1 mol S ___ 32 g

1,25 mol S __ 40 g

1 mol O₂ ___ 32 g

1,875 mol O₂ ___ 60 g

neutrons are in the middle of the atom and the electrons, which are much smaller.