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Balancear por el metodo oxidacion reduccion Mg + HNO3 → Mg(NO3)2 + N2 + H2O?

Balancear por el metodo oxidacion reduccion Mg + HNO3 → Mg(NO3)2 + N2 + H2O.

En resumen

₀ ₊₁ ₊₅ ₋₂ ₊₂ ₊₅ ₋₂ ₀ ₊₁ ₋₂Mg + H NO3 → Mg(NO3)2 + N2 + H2O1.

Mejor respuesta

Josueeljenio
10

₀ ₊₁ ₊₅ ₋₂ ₊₂ ₊₅ ₋₂ ₀ ₊₁ ₋₂Mg + H NO3 → Mg(NO3)2 + N2 + H2O1.

Semirreacciones ₀ ₊₂ Mg → Mg + 2 e⁻ (10 / 2 = 5 ) intercambiar electrones ₋₁ ₀ simplicar e⁻ 12H⁺ + 10e⁻ + 2(NO3) → N2 + 6H2O (2 / 2 = 1)`````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````` 5Mg⁰ + 12H⁺ + 2(NO3) ⁻¹ → 5Mg⁺² + N2⁰ + 6H2O 5 Mg + 12 HNO3 → 5Mg(NO3)2 + N2 + 6H2O.