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bija089 [108]
3 years ago
5

When 3.00 g of sulfur are combined with 3.00 g of oxygen, 6.00 g of sulfur dioxide (SO2) are formed. What mass of oxygen would b

e required to convert 6.00 g of sulfur into sulfur trioxide (SO3)?
Chemistry
1 answer:
bazaltina [42]3 years ago
4 0
Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.

2 S + 3 O₂ → 2 SO₃

The stoichiometric calculations is as follows:

6 g S * 1 mol/32.06 g S = 0.187 mol S
Moles O₂ needed = 0.187 mol S * 3 mol O₂/2 mol S = 0.2805 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.2805 mol O₂ * 32 g/mol = 8.976 g O₂
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<h3>Further explanation</h3>

Given

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Required

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Solution

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Cathode(reduction-negative pole) : 2H₂O(l)+2e⁻ ⇒ H₂(g)+2OH⁻(aq)

Anode(oxidation-positive pole) : 2H₂O(l)⇒O₂(g)+4H⁻(aq)+4e⁻

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So at the cathode H₂ gas is produced

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\tt mol~e^-=\dfrac{Q}{96500}

Q = i.t

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Q = 8100 C

mol e⁻ = 8100 : 96500 = 0.084

From equation at cathode , mol ratio e⁻ : H₂ = 2 : 1, so mol H₂ = 0.042

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