How many grams of sulfuric acid, H₂SO₄, can be obtained from 578 grams of iron ore if the ore is 76.0% by mass FeS? The reactions involved are given below. Each reaction is 92.0% efficient. 4 FeS + 7O₂ → 2 Fe₂O₃ + 4 SO₂
2 SO₂ + O₂ → 2 SO₃
SO₃ + H₂O → H₂SO₄

Question

Quizplus · Accepted Answer

First, calculate the mass of FeS in the ore: 578 g * 0.76 = 439.28 g FeS. The molar mass of FeS is 87.91 g/mol, so moles of FeS = 439.28 g / 87.91 g/mol = 5.00 mol. From the first reaction, 4 mol FeS produces 4 mol SO2, so 5.00 mol FeS produces 5.00 mol SO2. The efficiency is 92%, so actual moles of SO2 = 5.00 mol * 0.92 = 4.60 mol. From the second reaction, 2 mol SO2 produces 2 mol SO3, so 4.60 mol SO2 produces 4.60 mol SO3. The efficiency is 92%, so actual moles of SO3 = 4.60 mol * 0.92 = 4.23 mol. From the third reaction, 1 mol SO3 produces 1 mol H2SO4, so 4.23 mol SO3 produces 4.23 mol H2SO4. The molar mass of H2SO4 is 98.08 g/mol, so mass of H2SO4 = 4.23 mol * 98.08 g/mol = 414.58 g. The efficiency is 92%, so actual mass of H2SO4 = 414.58 g * 0.92 = 381.42 g, which rounds to 382 g.

How Many Grams of Sulfuric Acid, H2SO4, Can Be Obtained

How Many Grams of Sulfuric Acid, H₂SO₄, Can Be Obtained