Question

A)How many O2 molecules are needed to react with 7.80 g of S?

B)What is the theoretical yield of SO3 produced by the quantities described in Part A?

Answer 1

According to the question the reaction can be depicted as,

2S + 3O₂ → 2SO₃

A)

From the above equation we can observe that 2 moles sulphur (S) requires 3 moles of oxygen (O₂) for the reaction . Thus S and O₂ react in the ratio 2:3.

Mass of sulphur given = 7.80 g

No. of moles of Sulphur = Mass Given / Molar mass = 7.80 / 32 = 0.2438 mol          (Molar mass of sulphur = 32g)

So,

No. of moles of O₂ required = (0.2438) * (3/2) = 0.3657 mol

As we know,

No. of molecules of O₂ in 1 mole = 6.022 * 10²³ molecules (Avogadro Number)

No. of molecules of O₂ in 0.3657 mole = (6.022 * 10²³) * (0.3657) = 2.202 * 10²³ molecules

B)

The theoretical yield of SO₃ is the no. of moles of SO₃ produced in the reaction 2S + 3O₂ → 2SO₃ .

From the above equation we can clearly observe that 2 moles of SO₃ are produced for 2 moles of sulphur (S).

Thus, SO₃ and S react in the ratio 1:1

As we have calculated earlier,

Moles of Sulphur (S) present = 0.2438 mol

So,

No. of moles of SO₃ produced = 0.2438 mol

Molar Mass of SO₃ = Molar Mass of Sulphur + 3* (Molar Mass of oxygen) = 32 + 3*16 = 80 g

Mass of SO₃ produced = (No.of moles of SO₃) * (Molar mass of SO₃) = (0.2438) * (80 g) = 19.5 g

Thus, theoretical yield of SO₃ = 19.5 g