When reacting 760 grams of sulfur (S6) with an excess amount of oxygen (O2), you produce 1,190.0 g of sulfur dioxide (SO2). Calculate the theoretical yield and the percent yield.
S6 + 6O2 → 6SO2
chemistry?
2016-06-13 3:36 pm
回答 (1)
2016-06-13 4:14 pm
Molar mass of S₆ = 32.06×6 g/mol = 192.36 g/mol
Molar mass of SO₂ = 32.06 + 16.00×2 = 64.06 g/mol
S₆ + 6O₂ → 6SO₂
No. of moles of S₆ reacted = (760 g) / (192.36 g/mol) = 760/192.36 mol
Maximum no. of moles of SO₂ formed = (760/192.36 mol) × 6
Theoretical yield of SO₂ = (760/192.36 mol) × 6 × (64.06 g/mol) = 1518.6 g
Percent yield = (1190.0/1518.6) × 100% = 78.36%
Molar mass of SO₂ = 32.06 + 16.00×2 = 64.06 g/mol
S₆ + 6O₂ → 6SO₂
No. of moles of S₆ reacted = (760 g) / (192.36 g/mol) = 760/192.36 mol
Maximum no. of moles of SO₂ formed = (760/192.36 mol) × 6
Theoretical yield of SO₂ = (760/192.36 mol) × 6 × (64.06 g/mol) = 1518.6 g
Percent yield = (1190.0/1518.6) × 100% = 78.36%
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