Question

plesae ASAP

The contact process for producing sulfuric acid is: Sg(s) + 8 O2(g) — 2 SO2(g) + O2(g) SO3(g) + H2O(l) 8 SO2(g) +2 SO3(g) H2SO4(aq) each step has a percent yield of 77.1%, how many grams of Sg are needed to produce 1.50 kg sulfuric acid? 637 g 1.07*10^3 g 378 9 2.25*10^3 g 0 212 g

Answer 1

Correct answer is : 1.07*10^3 g

Explanation

mass H₂SO₄ = 1.50 kg = 1500 g

moles H₂SO₄ = (mass H₂SO₄) / (molar mass H₂SO₄)

moles H₂SO₄ = (1500 g) / (98.08 g/mol)

moles H₂SO₄ = 15.3 mol

theoretical moles H₂SO₄ = (actual moles H₂SO₄ produced) / (fraction yield)

theoretical moles H₂SO₄ = (15.3 mol) / (0.771)

theoretical moles H₂SO₄ = 19.8 mol

actual moles SO₃ consumed = theoretical moles H₂SO₄ produced

actual moles SO₃ consumed = 19.8 mol

theoretical moles SO₃ produced = (actual moles SO₃ produced) / (fraction yield)

theoretical moles SO₃ produced = (19.8 mol) / (0.771)

theoretical moles SO₃ produced = 25.7 mol

actual moles SO₂ consumed = theoretical moles SO₃ produced

actual moles SO₂ consumed = 25.7 mol

Theoretical moles SO₂ produced = (actual moles SO₂ consumed) / fraction yield)

Theoretical moles SO₂ produced = (25.7 mol) / (0.771)

Theoretical moles SO₂ produced = 33.4 mol

moles S₈ consumed = (Theoretical moles SO₂ produced) / 8

moles S₈ consumed = (33.4 mol) / 8

moles S₈ consumed = 4.17 mol

mass S₈ = (moles S₈) * (molar mass S₈)

mass S₈ = (4.17 mol) / (256.52 g/mol)

mass S₈ = 1070 g

mass S₈ = 1.7 x 10³ g