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Vinil7 [7]
3 years ago
14

2. Se combinan 40 g de SO2 y 25 g de O2 determine el porcentaje en masa del exceso con respecto a su masa inicial. P.A. (S = 32;

O = 16) SO2 + O2  SO3
Chemistry
1 answer:
Marta_Voda [28]3 years ago
8 0

Answer:

60%

Explanation:

En la reacción:

2SO₂ + O₂ → 2SO₃

<em>Donde 2 moles de SO₂ reaccionan por mol de O₂.</em>

El peso atómico del SO₂ es:

1S = 32g/molₓ1 = 32g/mol

2O = 16g/molₓ2 = 32g/mol

SO₂ = 32g/mol + 32g/mol = <em>64g/mol</em>

Y el peso atómico de O₂ es:

2O = 16g/molₓ2 = <em>32g/mol</em>

Las moles de SO₂ y O₂ son:

SO₂: 40g ₓ (1mol / 64g) = <em>0.625 moles</em>

O₂: 25g ₓ (1mol / 32g) = <em>0.7813 moles</em>

Para una completa reacción de SO₂ se necesitan:

0.625 moles SO₂ ₓ (1mol O₂ / 2 mol SO₂) = <em>0.3125 moles de O₂</em>

Así, las moles en exceso de O₂ son:

0.7813 moles - 0.3125 moles = 0.4688 moles O₂

En gramos:

0.4688 moles O₂ ₓ (32g / mol) = 15g O₂ están en exeso.

El porcentaje en exceso es:

15g / 25g ₓ 100 =

<h3>60%</h3>
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