Question

С какими из перечисленных веществ вступает в разные реакции раствор гидроксида натрия ;оксид бария,кальций,оксид серы(6),хлорид меди(2)?Составьте уравнения возможных реакций в молекулярном и ионном виде

Answer 1

2NaOH + SO3= Na2SO4 + H2O
CuCl2+2NaOH = Cu(OH)2↓+2NaCl
Cu²⁺+2Cl⁻+2Na⁺+2OH⁻ = Cu(OH)₂↓+ 2Na⁺+2Cl⁻
Cu²⁺+2OH⁻=Cu(OH)₂↓

Answer 2

Здравствуйте.

NaOH + BaO ≠ реакция не идёт, потому что основание не реагирует с основными оксидами.

NaOH + K ≠ реакция не идёт

2NaOH + SO₃ = Na₂SO₄ + H₂O

2Na⁺ + 2OH⁻ + SO₃ = 2Na⁺ + SO₄²⁻ + H₂O

2OH⁻ + SO₃ = SO₄²⁻ + H₂O

2NaOH + CuCl₂ = Cu(OH)₂ + 2NaCl

2Na⁺ + 2OH⁻ + Cu²⁺ + 2Cl⁻ = Cu(OH)₂ + 2Na⁺ + 2Cl⁻

Cu²⁺ + 2OH⁻ = Cu(OH)₂

Answer 3

Reactivity of Sodium Hydroxide Solution with Various Substances

Sodium hydroxide (NaOH) is a strong base and can react with various substances. Let's explore the reactions of sodium hydroxide solution with the listed substances: barium oxide, calcium, sulfur dioxide, and copper(II) chloride.

1. Barium Oxide (BaO) - Molecular Equation: ``` BaO + 2NaOH → Ba(OH)2 + Na2O ``` - Ionic Equation: ``` Ba^2+ + 2OH^- + 2Na^+ + 2OH^- → Ba^2+ + 2OH^- + 2Na^+ + O^2- ```

2. Calcium (Ca) - Molecular Equation: ``` Ca + 2NaOH → Ca(OH)2 + 2Na ``` - Ionic Equation: ``` Ca + 2OH^- + 2Na^+ + 2OH^- → Ca^2+ + 2OH^- + 2Na ```

3. Sulfur Dioxide (SO2) - Molecular Equation: ``` SO2 + 2NaOH → Na2SO3 + H2O ``` - Ionic Equation: ``` SO2 + 2Na^+ + 2OH^- → 2Na^+ + SO3^2- + H2O ```

4. Copper(II) Chloride (CuCl2) - Molecular Equation: ``` CuCl2 + 2NaOH → Cu(OH)2 + 2NaCl ``` - Ionic Equation: ``` Cu^2+ + 2Cl^- + 2Na^+ + 2OH^- → Cu(OH)2 + 2Na^+ + 2Cl^- ```

These reactions demonstrate the diverse reactivity of sodium hydroxide solution with different substances, resulting in the formation of various products.