balancear por el metodo oxidacion reduccion Mg + HNO3 → Mg(NO3)2 + N2 + H2O
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₀ ₊₁ ₊₅ ₋₂ ₊₂ ₊₅ ₋₂ ₀ ₊₁ ₋₂
Mg + H NO3 → Mg(NO3)2 + N2 + H2O
1. 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)
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5Mg⁰ + 12H⁺ + 2(NO3) ⁻¹ → 5Mg⁺² + N2⁰ + 6H2O
5 Mg + 12 HNO3 → 5Mg(NO3)2 + N2 + 6H2O
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