Question

Write a balanced net ionic equation for each of the following reactions: (a) Dilute nitric acid reacts with zinc metal with formation of nitrous oxide. (b) Concentrated nitric acid reacts with sulfur with formation of nitrogen dioxide. (c) Concentrated nitric acid oxidizes sulfur dioxide with formation of nitric oxide. (d) Hydrazine is burned in excess fluorine gas, forming \(\mathrm{NF}_{3}\). (e) \(\mathrm{Hy}\) drazine reduces \(\mathrm{CrO}_{4}{ }^{2-}\) to \(\mathrm{Cr}(\mathrm{OH})_{4}^{-}\) in base (hydrazine is oxidized to \(\mathrm{N}_{2}\) ).

Short answer

The balanced net ionic equations for each reaction are: a) Zn (s) + 2H⁺ (aq) → Zn²⁺ (aq) + N₂O (g) + H₂O (l) b) S (s) + 6H⁺ (aq) + 6NO₃⁻ (aq) → SO₄²⁻ (aq) + 6H₂O (l) + 6NO₂ (g) c) SO₂ (g) + 2H⁺ (aq) + 2NO₃⁻ (aq) → SO₄²⁻ (aq) + H₂O (l) + 2NO (g) d) N₂H₄ (g) + 3F₂ (g) → 2NF₃ (g) e) N₂H₄ (aq) + CrO₄²⁻ (aq) + 4OH⁻ (aq) → N₂ (g) + 2H₂O (l) + Cr(OH)₄⁻ (aq)

Step-by-step solution

a) Dilute nitric acid with zinc and nitrous oxide formation

1. Write the reactants and products in their ionic forms: Zn (s) + 2HNO₃ (aq) → Zn(NO₃)₂ (aq) + N₂O (g) + H₂O (l) 2. Balance the overall equation: Zn (s) + 2HNO₃ (aq) → Zn(NO₃)₂ (aq) + N₂O (g) + H₂O (l) 3. Write the net ionic equation by removing spectator ions: Zn (s) + 2H⁺ (aq) + NO₃⁻ (aq) → Zn⁺² (aq) + NO₃⁻ (aq) + N₂O (g) + H₂O (l) 4. Simplify the net ionic equation: Zn (s) + 2H⁺ (aq) → Zn²⁺ (aq) + N₂O (g) + H₂O (l)

b) Concentrated nitric acid with sulfur and nitrogen dioxide formation

1. Write the reactants and products in their ionic forms: 6HNO₃ (aq) + S (s) → H₂SO₄ (aq) + 6H₂O (l) + 6NO₂ (g) 2. Balance the overall equation: 6HNO₃ (aq) + S (s) → H₂SO₄ (aq) + 6H₂O (l) + 6NO₂ (g) 3. Write the net ionic equation by removing spectator ions: 6H⁺ (aq) + 6NO₃⁻ (aq) + S (s) → 2H⁺ (aq) + SO₄²⁻ (aq) + 6H₂O (l) + 6NO₂ (g) 4. Simplify the net ionic equation: S (s) + 6H⁺ (aq) + 6NO₃⁻ (aq) → SO₄²⁻ (aq) + 6H₂O (l) + 6NO₂ (g)

c) Concentrated nitric acid oxidizes sulfur dioxide with nitric oxide formation

1. Write the reactants and products in their ionic forms: 2HNO₃ (aq) + SO₂ (g) → H₂O (l) + SO₄²⁻ (aq) + 2NO (g) 2. Balance the overall equation: 2HNO₃ (aq) + SO₂ (g) → H₂O (l) + SO₄²⁻ (aq) + 2NO (g) 3. Write the net ionic equation by removing spectator ions: 2H⁺ (aq) + 2NO₃⁻ (aq) + SO₂ (g) → H₂O (l) + SO₄²⁻ (aq) + 2NO (g) 4. Simplify the net ionic equation: SO₂ (g) + 2H⁺ (aq) + 2NO₃⁻ (aq) → SO₄²⁻ (aq) + H₂O (l) + 2NO (g)

d) Hydrazine burned in excess fluorine gas forming NF₃

1. Write the reactants and products in their ionic forms: N₂H₄ (g) + 3F₂ (g) → 2NF₃ (g) 2. Balance the overall equation: N₂H₄ (g) + 3F₂ (g) → 2NF₃ (g) 3. No ions are involved in this reaction, so the net ionic equation is the same as the balanced overall equation: N₂H₄ (g) + 3F₂ (g) → 2NF₃ (g)

e) Hydrazine reduces CrO₄²⁻ to Cr(OH)₄⁻ in base, with hydrazine oxidized to N₂

1. Write the reactants and products in their ionic forms: N₂H₄ (aq) + CrO₄²⁻ (aq) + 4OH⁻ (aq) → N₂ (g) + 2H₂O (l) + Cr(OH)₄⁻ (aq) 2. Balance the overall equation: N₂H₄ (aq) + CrO₄²⁻ (aq) + 4OH⁻ (aq) → N₂ (g) + 2H₂O (l) + Cr(OH)₄⁻ (aq) 3. Write the net ionic equation by removing spectator ions (none in this case, as all ions are involved in the reaction): N₂H₄ (aq) + CrO₄²⁻ (aq) + 4OH⁻ (aq) → N₂ (g) + 2H₂O (l) + Cr(OH)₄⁻ (aq) 4. Simplify the net ionic equation (no simplification necessary as all ions are involved in the reaction): N₂H₄ (aq) + CrO₄²⁻ (aq) + 4OH⁻ (aq) → N₂ (g) + 2H₂O (l) + Cr(OH)₄⁻ (aq)