Answer to Question #269711 in General Chemistry for nedo

Question

. How many grams of sulfuric acid, H2SO4, can be obtained from 457 grams of iron ore if the ore is 73.0% by mass FeS? The reactions involved are given below. Reactions happen with the yield of 90.0%, 82.0% and 92.5% respectively 4 FeS + 7O2 → 2 Fe2O3 + 4 SO2 2 SO2 + O2 → 2 SO3 SO3 + H2O → H2SO4 

Answer 1

mass of iron ore = 457g

mass of FeS in iron ore = 73.0% × 457g = 333.6g

number of moles of FeS = 333.6/87.91 = 3.8 moles

According to the reaction above,

First reaction,

4 moles of FeS produces 4 moles of SO₂

$\therefore$ 3.8 moles of SO₂ are produced

but the yield of SO₂ is 90%

The actual number of moles of SO₂ produced = 90.0% × 3.8mol = 3.42mol

In the second reaction,

4 moles of SO₂ produces 4 moles of SO₃,

This means that 3.42 moles of SO₃ is produced.

The actual number of moles of SO₃ produced is 82.0% × 3.42 moles = 2.804 moles

In the third reaction,

4 moles of SO₃ produces 4 moles of H₂SO₄.

This means that 2.804 moles of H₂SO₄.

The actual number of moles produced is 92.5% of 2.804 moles = 2.6 moles

The total mass of H₂SO₄ produced is 2.6 moles × 98g/mol = $255\textsf {g}.$