Answer to Question #152551 in Chemistry for Andrew Liu

Question

Determine the volume of SO2 (at STP) formed from the reaction of 96.7 g of FeS2 and 55.0 L of O2 (at 398 K and 1.20 bar). The molar mass of FeS2 is 119.99 g mol-1.

4FeS2(s) + 11O2(g) → 2Fe2O3(s) + 8SO2(g)

Answer 1

R = 8.314×10⁻² L⋅bar⋅K⁻¹⋅mol⁻¹ (universal gas constant).

Solution:

The balanced chemical equation:

4FeS₂(s) + 11O₂(g) → 2Fe₂O₃(s) + 8SO₂(g)

According to the equation: n(FeS₂)/4 = n(O₂)/11 = n(SO₂)/8

1) Moles of FeS₂ = Mass of FeS₂ / Molar mass of FeS₂

The molar mass of FeS₂ is 119.99 g mol^-1.

Hence,

n(FeS₂) = Moles of FeS₂ = 96.7 g / 119.99 g mol^-1 = 0.806 mol

n(FeS₂) = 0.806 mol

2) According to the ideal gas law: PV = nRT

Moles of O₂ = PV / RT

n(O₂) = Moles of O₂ = (1.20 bar × 55.0 L) / (8.314×10⁻² L⋅bar⋅K⁻¹⋅mol⁻¹ × 398 K ) = 1.9946 mol

n(O₂) = 1.9946 mol

According to stoichiometry:

11 moles of O₂ reacts with 4 moles of FeS₂

Thus 1.9946 moles of oxygen reacts with:

(1.9946 mol O₂ × 4 mol FeS₂) / 11 mol O₂ = 0.7253 mol FeS₂.

However, initially there is 0.806 mol of FeS₂ (according to the task).

Thus O₂ acts as limiting reagent and FeS₂ is excess reagent.

Therefore,

n(O₂)/11 = n(SO₂)/8

n(SO₂) = 8 × n(O₂) / 11 = (8 × 1.9946 mol) / 11 = 1.4506 mol SO₂

n(SO₂) = 1.4506 mol

At STP, one mole of any gas occupies a volume of 22.4 L.

Thus 1.4506 moles of SO₂ occupies:

(1.4506 mol × 22.4 L) / 1 mol = 32.49 L of SO₂

V(SO₂) = 32.49 L = 32.5 L (at STP).

Answer: 32.5 L of SO₂ are formed.