Question

Составьте уравнения реакций взаимодействия ZnCl2 и NaOH Молекулярное уравнение: Полное ионное уравнение: Сокращенное ионное уравнение: Реакция ионного обмена идет до конца и почему?

Answer 1

Ответ:

Объяснение:

ZnCl2 + 2NaOH = Zn(OH)2↓ + 2NaCl

Zn(2+) + 2Cl(-) + 2Na(+) + 2OH(-) = Zn(OH)2↓ + 2Na(+) + 2Cl(-)

Zn(2+) + 2OH(-) = Zn(OH)2↓

Реакция идёт до конца - выделяется осадок

Answer 2

Molecular Equation:

The molecular equation for the reaction between ZnCl2 and NaOH is as follows:

ZnCl2 + 2NaOH -> Zn(OH)2 + 2NaCl

Complete Ionic Equation:

The complete ionic equation for the reaction between ZnCl2 and NaOH is as follows:

Zn^2+(aq) + 2Cl^-(aq) + 2Na^+(aq) + 2OH^-(aq) -> Zn(OH)2(s) + 2Na^+(aq) + 2Cl^-(aq)

Net Ionic Equation:

The net ionic equation for the reaction between ZnCl2 and NaOH is as follows:

Zn^2+(aq) + 2OH^-(aq) -> Zn(OH)2(s)

Explanation:

The reaction between ZnCl2 and NaOH is an example of an ion exchange reaction. In this reaction, the zinc cations (Zn^2+) from ZnCl2 react with the hydroxide anions (OH^-) from NaOH to form zinc hydroxide (Zn(OH)2) as a solid precipitate. The sodium cations (Na^+) and chloride anions (Cl^-) remain in the solution as spectator ions.

The reaction proceeds to completion because the formation of a solid precipitate (Zn(OH)2) reduces the concentration of the hydroxide ions (OH^-) in the solution. This shift in equilibrium drives the reaction forward until all the available hydroxide ions have reacted with the zinc cations to form the precipitate.