fecl3 + h2s = fecl2 +s +hcl
hno3 +i = no2 + hio3 + h2o como quedaría este ejercicio con el método de oxido- Redox .explique porfa paso a paso
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₊₃ ₋₁ ₊₁ ₋₂ ₊₂ ₋₁ ₀ ₊₁ ₋₂
2FeCl3 + H2 S → 2FeCl2 + S + 2HCl
1e⁻ + Fe⁺³ → Fe⁺² Ι 2
H2S → S + 2H⁺ + 2e⁻ Ι 1 intercambiar electrones
````````````````````````````````````````````
2e⁻ + 2Fe⁺³ → Fe⁺²
H2S → S + 2H⁺ + 2e⁻ simplificar electrones
````````````````````````````````````````````````
2Fe⁺³ + H2S → Fe⁺² + S + 2H⁺
````````````````````````````````````````````````````````````````````````````````````````````````````
₊₂₊₅₋₂ ₀ ₊₄₋₂ ₊₁₊₅₋₂ ₊₁₋₂
5 HNO3 + I →5 NO2 + H I O3 + 2 H2O
₀
3H2O + I → (IO3)⁻¹ + 6H⁺ + 5e⁻ Ι`1
2H⁺ + 1e⁻ + (NO3)⁻¹ → NO2 + H2O Ι 5
````````````````````````````````````````````````````````````````````````````````````
₀
3H2O + I + 10H⁺ + 5e⁻ + 5(NO3)⁻¹ → (IO3)⁻¹ + 6H⁺ + 5e⁻ + 5NO2 + 5H2O
SIMPLIFICAR LAS MOLECLAS DE AGUA, H⁺ y e⁻
₀
4H⁺ + I + 5(NO3)⁻¹ → (IO3)⁻¹ + 5NO2 + 2H2O
2FeCl3 + H2 S → 2FeCl2 + S + 2HCl
1e⁻ + Fe⁺³ → Fe⁺² Ι 2
H2S → S + 2H⁺ + 2e⁻ Ι 1 intercambiar electrones
````````````````````````````````````````````
2e⁻ + 2Fe⁺³ → Fe⁺²
H2S → S + 2H⁺ + 2e⁻ simplificar electrones
````````````````````````````````````````````````
2Fe⁺³ + H2S → Fe⁺² + S + 2H⁺
````````````````````````````````````````````````````````````````````````````````````````````````````
₊₂₊₅₋₂ ₀ ₊₄₋₂ ₊₁₊₅₋₂ ₊₁₋₂
5 HNO3 + I →5 NO2 + H I O3 + 2 H2O
₀
3H2O + I → (IO3)⁻¹ + 6H⁺ + 5e⁻ Ι`1
2H⁺ + 1e⁻ + (NO3)⁻¹ → NO2 + H2O Ι 5
````````````````````````````````````````````````````````````````````````````````````
₀
3H2O + I + 10H⁺ + 5e⁻ + 5(NO3)⁻¹ → (IO3)⁻¹ + 6H⁺ + 5e⁻ + 5NO2 + 5H2O
SIMPLIFICAR LAS MOLECLAS DE AGUA, H⁺ y e⁻
₀
4H⁺ + I + 5(NO3)⁻¹ → (IO3)⁻¹ + 5NO2 + 2H2O
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