Question

Consider the following unbalanced reaction.

SO₂(g) + O₂(g) → SO₃(g)

How many grams of O₂ are needed to produce 115 g of SO₃ if the reaction has a 77.5% yield?

Answer 1

The balanced equation is

2 SO₂(g) + O₂(g) → 2 SO₃(g)

Number of moles of SO3 = 115 g / 80.066 g/mol = 1.44 mole

From the balanced equation we can say that

2 mole of SO3 requires 1 mole of O2 so

1.44 mole of SO3 will require

= 1.44 mole of SO3 *(1 mole of O2 / 2 mole of SO3)

= 0.720 mole of O2

mass of 1 mole of O2 = 32.0 g so

the mass of 0.720 mole of O2 = 23.0 g

Therefore, the mass of O2 required would be 23.0 g