Chemistry: half reaction problem

Question

How much of H₂SO₄ will be produced after oxydizing 2 kg of pyrite FeS₂, which has mass part of iron sulfide of 95.2%.

Subsequently, how many litres of 5 mol/l KOH do we need to neutralize the resulting H₂SO₄.

The unbalanced reaction equation:

FeS₂ (solid) + O₂ (gas) + H₂O + ... -> FeSO₄ (aq) + H₂SO₄ (aq)

Answer 1

2 FeS₂ (solid) + 7 O₂ (gas) + 2 H₂O ==>2 FeSO₄ (aq) + 2 H₂SO₄ (aq) .. balanced equation

molar mass FeS₂ = 120 g / mole

2 kg x 1000 g / kg x 95.2% FeS2 = 1904 g FeS₂

moles FeS₂ = 1904 g FeS₂ x 1 mol / 120 g = 15.87 moles

moles H₂SO₄ produced = 15.87 moles FeS₂ x 2 mol H2_SO₄ / 2 mol FeS₂ = 15.87 mols H₂SO₄

To neutralize this with 5 M KOH:

H₂SO₄ + 2KOH ==> 2H₂O + K₂SO₄

15.87 mol H₂SO₄ x 2 mol KOH / mol H₂SO₄ = 31.74 mols KOH

31.74 mols KOH x 1 L / 5 mols = 6.35 L KOH