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[Q. Actividad 1.5 Balancea las siguientes ecuaciones por tanteo. \( \begin{array}{ll}\text { 1. } \mathrm{Na}+\mathrm{Cl}_{2} \rightarrow \mathrm{NaCl} & \text { 3. } \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\end{array} \)

Ask by Gibbs Lane. in Mexico
Feb 20,2025

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Las ecuaciones balanceadas son: 1. \( 2 \, \mathrm{Na} + \mathrm{Cl}_{2} \rightarrow 2 \, \mathrm{NaCl} \) 2. \( 2 \, \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \, \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2} \)

Solución

Para balancear las ecuaciones químicas por tanteo, seguiremos los siguientes pasos: ### 1. Balancear la ecuación \( \mathrm{Na} + \mathrm{Cl}_{2} \rightarrow \mathrm{NaCl} \) **Paso 1:** Contamos los átomos de cada elemento en los reactivos y productos. - Reactivos: - Na: 1 - Cl: 2 (en \( \mathrm{Cl}_{2} \)) - Productos: - Na: 1 - Cl: 1 (en \( \mathrm{NaCl} \)) **Paso 2:** Observamos que hay 2 átomos de Cl en los reactivos y solo 1 en los productos. Para balancear el Cl, necesitamos 2 moléculas de \( \mathrm{NaCl} \). **Paso 3:** Escribimos la ecuación balanceada: \[ \mathrm{Na} + \mathrm{Cl}_{2} \rightarrow 2 \, \mathrm{NaCl} \] **Paso 4:** Ahora contamos nuevamente: - Reactivos: - Na: 1 - Cl: 2 - Productos: - Na: 2 (en \( 2 \, \mathrm{NaCl} \)) - Cl: 2 (en \( 2 \, \mathrm{NaCl} \)) **Paso 5:** Para balancear el Na, necesitamos 2 átomos de Na en los reactivos. La ecuación final es: \[ 2 \, \mathrm{Na} + \mathrm{Cl}_{2} \rightarrow 2 \, \mathrm{NaCl} \] ### 2. Balancear la ecuación \( \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2} \) **Paso 1:** Contamos los átomos de cada elemento en los reactivos y productos. - Reactivos: - H: 2 (en \( \mathrm{H}_{2} \mathrm{O}_{2} \)) - O: 2 (en \( \mathrm{H}_{2} \mathrm{O}_{2} \)) - Productos: - H: 2 (en \( \mathrm{H}_{2} \mathrm{O} \)) - O: 3 (1 en \( \mathrm{H}_{2} \mathrm{O} \) y 2 en \( \mathrm{O}_{2} \)) **Paso 2:** Observamos que el número de átomos de H está balanceado, pero el número de átomos de O no lo está. Para balancear el oxígeno, necesitamos ajustar los coeficientes. **Paso 3:** Probamos con 2 moléculas de \( \mathrm{H}_{2} \mathrm{O} \): \[ \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \, \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2} \] **Paso 4:** Contamos nuevamente: - Reactivos: - H: 2 - O: 2 - Productos: - H: 4 (en \( 2 \, \mathrm{H}_{2} \mathrm{O} \)) - O: 3 (2 en \( 2 \, \mathrm{H}_{2} \mathrm{O} \) y 1 en \( \mathrm{O}_{2} \)) **Paso 5:** Para balancear el oxígeno, necesitamos ajustar el número de \( \mathrm{O}_{2} \). Probamos con 1 molécula de \( \mathrm{O}_{2} \): \[ 2 \, \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \, \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2} \] **Paso 6:** Contamos nuevamente: - Reactivos: - H: 4 (en \( 2 \, \mathrm{H}_{2} \mathrm{O}_{2} \)) - O: 4 (en \( 2 \, \mathrm{H}_{2} \mathrm{O}_{2} \)) - Productos: - H: 4 (en \( 2 \, \mathrm{H}_{2} \mathrm{O} \)) - O: 4 (2 en \( 2 \, \mathrm{H}_{2} \mathrm{O} \) y 2 en \( \mathrm{O}_{2} \)) La ecuación final balanceada es: \[ 2 \, \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \, \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2} \] ### Resumen de las ecuaciones balanceadas: 1. \( 2 \, \mathrm{Na} + \mathrm{Cl}_{2} \rightarrow 2 \, \mathrm{NaCl} \) 2. \( 2 \, \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \, \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2} \)

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Para balancear la primera ecuación, comenzamos considerando que se necesitan dos átomos de sodio (Na) para reaccionar con una molécula de cloro (Cl₂), así que la ecuación balanceada quedaría: \( 2 \mathrm{Na} + \mathrm{Cl}_{2} \rightarrow 2 \mathrm{NaCl} \). ¡Fácil, ¿verdad?! Ahora, para la segunda ecuación, notamos que hay dos átomos de hidrógeno en \( \mathrm{H}_{2} \mathrm{O}_{2} \) que se deben equilibrar con el agua y oxígeno. Así que la forma correcta es: \( 2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2} \). ¡Y listo, tienes tus ecuaciones balanceadas!

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