If 7.99 g of sulfur dioxide reacts with 2.18 g of oxygen, how many grams of sulfur trioxide can form?

Question

Another way that sulfuric acid is formed in the atmosphere is when sulfur dioxide reacts with oxygen in a reaction catalyzed by dust in the atmosphere to form sulfur trioxide. If 7.99 g of sulfur dioxide reacts with 2.18 g of oxygen, how many grams of sulfur trioxide can form?

Answer 1

Limiting reactant problem

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

One way to do this is to simply divide the moles of each reactant by the corresponding coefficient in the balanced equation.

2SO₂ + O₂ ==> 2SO₃ ... BALANCED EQUATION

For SO₂: 7.99 g SO₂ x 1 mol SO₂ / 64.1 g = 0.1246 mols SO₂ (÷2->0.0623)

For O₂: 2.18 g O₂ x 1 mol O₂ / 32 g = 0.0681 mols O₂ (÷1->0.0681)

Since 0.0623 is less than 0.0681, SO₂ is the limiting reactant (not by much, but it is).

Because SO₂ is limiting it will determine how much product (SO₃) can be formed

0.1246 mols SO₂ x 2 mols SO₃ / 2 mols SO₂ x 80.1 g SO₃ / mols SO3 = 9.98 g SO₃