Question

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?

Answer 1

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

Step-by-step 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