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faltersainse [42]
3 years ago
10

Alguien me puede hacer el favor de ayudarme con estos 3 ejercicios de balanceo de ecuaciones por oxido reducción por favor 1. Fe

2O3 + CO → Fe + CO2 2. Al2O3 + C + Cl2 → CO + AlCl3 3. H2S + O2 → SO2 + H2O
Chemistry
1 answer:
kondaur [170]3 years ago
5 0

Answer:

1. Fe₂O₃ + 3CO → 2Fe + 3CO₂

2. Al₂O₃ + 3C + Cl₂ → 3CO + 2AlCl₃

3. 2H₂S + 3O₂ → 2SO₂ + 2H₂O

Explanation:

Para balancear ecuaciones redox se acostumbra a usar el método ion-electrón, hay que determinar qué elemento se oxida y cual se reduce lo que a la vez significa, aumento del número de oxidación o descenso del mismo.

1. Fe₂O₃ + CO → Fe + CO₂

Para este caso, el hierro se reduce, pasando de Fe³⁺ a Fe⁰ o lo que se dice estado fundamental.

Fe³⁺  +  3e⁻  → Fe

El carbono del CO se oxida a CO₂ donde el carbono pasa de C²⁺ a C⁴⁺

C²⁺  →   C⁴⁺ + 2e⁻

Como los electrones quedaron desbalanceados multiplicamos por 2 y 3 las semi reacciones

(Fe³⁺  +  3e⁻  → Fe ) . 2

2Fe³⁺  +  6e⁻  → 2Fe

(C²⁺  →   C⁴⁺ + 2e⁻) . 3

3C²⁺  →   3C⁴⁺ + 6e⁻

Ya podemos balancear la ecuacion final:

Fe₂O₃ + 3CO → 2Fe + 3CO₂

(como ya tenemos 2 atomos de hierro en el oxido, no haría falta agregar el 2 en la estoquiometría)

2. Al₂O₃ + C + Cl₂ → CO + AlCl₃

En este caso el carbono se oxida, y el cloro se reduce

C →  C²⁺  +  2e⁻

Cl₂ + 2e⁻ → 2Cl⁻

Como el cloro es una molécula diatómica por cada cloro que se reduce, tendremos dos cloruros, tomando los 2 electrones que libera el C.

En este caso los electrones están balanceados

Al₂O₃ + C + Cl₂ → CO + 2AlCl₃

Han quedado desbalanceados el C y el O, asi que no queda otra que completar con la estequiometría

Al₂O₃ + 3C + Cl₂ → 3CO + 2AlCl₃

3.  H₂S + O₂ → SO₂ + H₂O

El O₂ se está reduciendo, pasando de 0 a -2

El S del ácido se oxida de +2 a +4

O₂ + 4e⁻ → 2O⁻²

Son dos atomos de O, que cada átomo puede atrapar dos electrones, por ende, consigue enlazar 4 electrones en total

S⁺²  → S⁴⁺ + 2e⁻

Como los electrones quedaron desbalanceados, completamos multiplicando la segunda semi reaccion x2

2S⁺²  → 2S⁴⁺ + 4e⁻

O₂ + 4e⁻ → 2O⁻²

2H₂S + O₂ → 2SO₂ + 2H₂O

Ya que el oxigeno quedó desbalanceado en el lado producto, completamos con la estoquimetría

2H₂S + 3O₂ → 2SO₂ + 2H₂O

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After calculating forΔG we found that the value ofΔG is negative and its value is -101.74KJ

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As the ΔG for the given reaction is is negative so the reaction will be spontaneous in nature.

In this reaction since the entropy of reaction is positive and hence when we increase the temperature term then the overall term TΔS would become more positive  and hence the value of ΔG would be less negative .

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vekshin1

<u>Answer:</u> The final temperature of the copper is 95°C.

<u>Explanation:</u>

To calculate the final temperature for the given amount of heat absorbed, we use the equation:

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