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Answer :
When Methane thiol, CH₃SH, reacts with O₂, 25.8 grams of Sulfur dioxide, SO₂ is formed and limiting reactant is Oxygen, O₂.
The balanced chemical reaction is:
CH₃SH(g) + 3 O₂(g) → 2 H₂O(g) + CO₂(g) + SO₂(g)
Molecular weights: Actual/given mass:
CH₃SH = 48.11 g/mol 27.3 g
O₂ = 32.01 g/mol 38.6 g
H₂O = 18.01 g/mol
CO₂ = 44.01 g/mol
SO₂ = 64.06 g/mol
Number of moles = Given mass
Molecular mass
So,
moles of methane thiol = 27.3 ÷ 48.11= 0.56 moles
moles of oxygen = 32.01÷38.6 = 0.82 moles
From the reaction,
1 mole of CH₃SH reacts with 3 moles of O₂ to give 1 mole of SO₂
Thus, 0.56 moles of CH₃SH reacts to form 1 × 0.56 = 0.56 moles of SO₂
Mass of SO₂ produced is 0.56 × 64.06 g = 35.86 g
moles of SO₂ = 35.86 ÷ 64.06 = 0.55 moles
So,
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