Sulfur dioxide gas (SO2) and oxygen gas (O2) react to form the liquid product of sulfur trioxide (SO3).
Express the amount in grams to three significant digits.
How much sulfur dioxide would you need to completely react with 6.00 g O2 such that all reactants could be consumed?
2017-10-20 10:36 am
回答 (1)
2017-10-20 10:43 am
Molar mass of SO₂ = (32.1 + 16.0×2) g/mol = 64.1 g/mol
Molar mass of O₂ = 16.0 × 2 g/mol = 32.0 g/mol
2SO₂(g) + O₂(g) → 2SO₃(l)
Mole ratio SO₂ : O₂ = 2 : 1
No. of moles of O₂ reacted = (6.00 g) / (32.0 g/mol) = 0.1875 mol
No. of moles of SO₂ needed = (0.1875 mol) × 2 = 0.375 mol
Mass of SO₂ needed = (0.375 mol) × (64.1 g/mol) = 24.0 g
Molar mass of O₂ = 16.0 × 2 g/mol = 32.0 g/mol
2SO₂(g) + O₂(g) → 2SO₃(l)
Mole ratio SO₂ : O₂ = 2 : 1
No. of moles of O₂ reacted = (6.00 g) / (32.0 g/mol) = 0.1875 mol
No. of moles of SO₂ needed = (0.1875 mol) × 2 = 0.375 mol
Mass of SO₂ needed = (0.375 mol) × (64.1 g/mol) = 24.0 g
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