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. According to the first reaction, 4 mol FeS produces 4 mol SO2, so 5.00 mol FeS produces 5.00 mol SO2. The second reaction shows 2 mol SO2 produces 2 mol SO3, so 5.00 mol SO2 produces 5.00 mol SO3. The third reaction shows 1 mol SO3 produces 1 mol H2SO4, so 5.00 mol SO3 produces 5.00 mol H2SO4. The molar mass of H2SO4 is 98.08 g/mol, so mass of H2SO4 = 5.00 mol * 98.08 g/mol = 490.4 g. Considering 92% efficiency for each step: 490.4 g * 0.92 * 0.92 * 0.92 = 382 g.

How Many Grams of Sulfuric Acid,H2SO4,can Be Obtained from 578

How Many Grams of Sulfuric Acid,H₂SO₄,can Be Obtained from 578