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Zolol [24]
3 years ago
15

Se queman 87,0 g de butano con oxígeno produciéndose dióxido de carbono y agua. Calcular la masa de oxígeno necesaria y la masa

de dióxido de carbono y de agua que se desprenderá
Chemistry
1 answer:
galina1969 [7]3 years ago
4 0

Answer:

312g O₂ son necesarios, se producen 264g CO₂ y 135g H₂O

Explanation:

La combustión de butano para producir CO₂ y H₂O ocurre así:

C₄H₁₀ + 13/2O₂ → 4CO₂ + 5H₂O

<em>Donde 1 mol de butano reacciona con 13 moles de oxígeno para producir 4 moles de dióxido de carbono y 5 moles de agua.</em>

<em />

Para resolver este problema debemos convertir la masa de butano a moles y, usando la reacción balanceada, obtener cuánto se necesita de oxígeno y cuánto se produciría de dióxido de carbono y agua:

<em>Moles butano -Masa molar: 58.12g/mol-:</em>

87.0g * (1mol / 58.12g) = 1.50 moles

<em>Moles y masa de oxígeno necesarias para la reacción -Masa molar: 32g/mol-:</em>

1.50 moles C₄H₁₀ * (13/2 moles O₂ / 1 mol C₄H₁₀) = 9.75 moles * (32g/mol) =

312g O₂ son necesarios

<em>Moles y masa de dióxido de carbono producidas -Masa molar: 44.01g/mol-:</em>

1.50 moles C₄H₁₀ * (4 moles CO₂ / 1 mol C₄H₁₀) = 6 moles * (44.01g/mol) =

264g CO₂ se producen

<em>Moles y masa de agua producidas -Masa molar: 18.015g/mol-:</em>

1.50 moles C₄H₁₀ * (5 moles H₂O / 1 mol C₄H₁₀) = 7.5 moles * (18.015g/mol) =

135g H₂O se producen

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