Question

1. Sulfuric acid can be produced by the following sequence of reactions.

             S₈ + 8 O₂ --> 8 SO₂

              2 SO₂ + O₂   --> 2 SO₃

              SO₃ + H₂O --> H₂SO₄

You find a source of sulfur that is 65.0% sulfur by mass. How many grams of this starting material would be required to produce 87.0 g of sulfuric acid? Explain and show work.

Answer 1

Moles of sulfuric acid required = 87.0g/98.0785g/mol = 0.887045moles

Moles of SO₃ required = moles of sulfuric acid = 0.887045

Moles of SO₂ required = (2/2)×moles of SO₃ = 0.887045

Moles of S₈ = (1/8)×moles of SO₂ = (1/8)×0.887045

= 0.11088mol

Mass of sulfur required = number of moles × molar mass = 0.11088mol × 256.52g/mol = 28.4431 g

Amount of starting material required = (100/65)×28.4431g

= 43.7586 g

= 43.8 g. (Answer)