Respuesta :
The net ionic equation of chromium (iii) hydroxide reacting with nitrous acid:
Cr(OH)₃ (s) + 3H⁺ (aq) ⇒ Cr³⁺ (aq) + 3H₂O (l)
Further explanation
The electrolyte in the solution produces ions.
The equation of a chemical reaction can be expressed in the equation of the ions
For strong electrolytes (the ionization rate = 1) is written in the form of separate ions, while the weak electrolyte (degree of ionization <1) is still written as an un-ionized molecule
In the ion equation, there is a spectator ion that is the ion which does not change (does not react) because it is present before and after the reaction
When these ions are removed, the ionic equation is called the net ionic equation
For gases and solids including water (H₂O) can be written as an ionized molecule
So only the dissolved compound is ionized ((expressed in symbol aq)
From the problem, it is stated that chromium (iii) hydroxide reacting with nitrous acid.
Reactions that occur:
Cr(OH)₃ (s) + 3HNO₂ (aq) ⇒ Cr(NO₂)₃ (aq) + 3H₂O (l)
Cr (OH)₃ solid form that does not decompose in the form of ions, as well as water (H₂O)
So the ionic equation becomes:
Cr(OH)₃ (s) + 3H⁺ (aq) + 3NO₂⁻ (aq) ⇒ Cr³⁺ (aq) + 3NO₂⁻ (aq) + 3H₂O (l)
There is an ion spectator that is 3NO₂⁻, so that if it is removed a net ionic equation will be formed:
Cr(OH)₃ (s) + 3H⁺ (aq) ⇒ Cr³⁺ (aq) + 3H₂O (l)
Learn more
the net ionic equation
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Keywords: the net ionic equation, an ion spectator
The net ionic equation is [tex]\boxed{{\text{Cr}}{{\left( {{\text{OH}}} \right)}_3}\left( s \right)+3{{\text{H}}^+}\left( {aq} \right)\to{\text{C}}{{\text{r}}^{3+}}\left( {aq}\right)+3{{\text{H}}_{\text{2}}}{\text{O}}\left(l\right)}[/tex].
The phase of [tex]{\mathbf{Cr}}{\left( {{\mathbf{OH}}} \right)_{\mathbf{3}}}[/tex] is the solid phase, that of [tex]{{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}[/tex] is liquid while that of [tex]{{\mathbf{H}}^{\mathbf{ + }}}[/tex] and [tex]{\mathbf{C}}{{\mathbf{r}}^{{\mathbf{3 + }}}}[/tex] is aqueous.
Further Explanation:
The three types of equations that are used to represent the chemical reaction are as follows:
1. Molecular equation
2. Total ionic equation
3. Net ionic equation
The reactants and products remain in undissociated form in molecular equation. In the case of total ionic equation, all the ions that are dissociated and present in the reaction mixture are represented while in the case of overall or net ionic equation only the useful ions that participate in the reaction are represented.
The steps to write the net ionic reaction are as follows:
Step 1: Write the molecular equation for the reaction with the phases in the bracket.
In the reaction, [tex]{\text{Cr}}{\left( {{\text{OH}}} \right)_3}[/tex] reacts with [tex]{\text{HN}}{{\text{O}}_2}[/tex] to form [tex]{\text{Cr}}{\left( {{\text{N}}{{\text{O}}_{\text{2}}}} \right)_3}[/tex] and [tex]{{\text{H}}_2}{\text{O}}[/tex]. The balanced molecular equation of the reaction is as follows:
[tex]{\text{Cr}}{\left( {{\text{OH}}} \right)_3}\left( s \right) + {\text{3HN}}{{\text{O}}_2}\left( {aq} \right) \to {\text{Cr}}{\left( {{\text{N}}{{\text{O}}_2}} \right)_3}\left( {aq} \right) + {\text{3}}{{\text{H}}_2}{\text{O}}\left( l \right)[/tex]
Step 2: Dissociate all the compounds with the aqueous phase to write the total ionic equation. The compounds with liquid and solid phase remain the same. The total ionic equation is as follows:
[tex]{\text{Cr}}{\left( {{\text{OH}}} \right)_3}\left( s \right) + {\text{3}}{{\text{H}}^ + }\left( {aq} \right) + {\text{3NO}}_2^ - \left( {aq} \right) \to 3{\text{C}}{{\text{r}}^{3 + }}\left( {aq} \right) + {\text{3NO}}_2^ - \left( {aq} \right) + {\text{3}}{{\text{H}}_2}{\text{O}}\left( l \right)[/tex]
Step 3: The common ions on both the sides of the reaction get cancelled out to get the overall ionic equation.
[tex]{\text{Cr}}{\left( {{\text{OH}}} \right)_3}\left( s \right) + {\text{3}}{{\text{H}}^ + }\left( {aq} \right) + \boxed{{\text{3NO}}_2^ - \left( {aq} \right)} \to 3{\text{C}}{{\text{r}}^{3 + }}\left( {aq} \right) + \boxed{{\text{3NO}}_2^ - \left( {aq} \right)} + {\text{3}}{{\text{H}}_2}{\text{O}}\left( l \right)[/tex]
Therefore, the overall ionic equation obtained is as follows:
[tex]{\text{Cr}}{\left( {{\text{OH}}} \right)_3}\left( s \right) + {\text{3}}{{\text{H}}^ + }\left( {aq} \right) \to 3{\text{C}}{{\text{r}}^{3 + }}\left( {aq} \right) + {\text{3}}{{\text{H}}_2}{\text{O}}\left( l \right)[/tex]
[tex]{\mathbf{Cr}}{\left( {{\mathbf{OH}}} \right)_{\mathbf{3}}}[/tex] is present in the solid phase, [tex]{{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}[/tex] is in the liquid phase while [tex]{{\mathbf{H}}^{\mathbf{ + }}}[/tex] and [tex]{\mathbf{C}}{{\mathbf{r}}^{{\mathbf{3 + }}}}[/tex] are present in the aqueous phase.
Learn more:
1. Balanced chemical equation: https://brainly.com/question/1405182
2. The net ionic equation for the reaction of [tex]{\text{MgS}}{{\text{O}}_{\text{4}}}[/tex] with [tex]{\text{Sr}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}[/tex]: https://brainly.com/question/4357519
Answer details:
Grade: High School
Subject: Chemistry
Chapter: Chemical reaction and equation
Keywords: net ionic equation, H2O, Cr(OH)3, H+, Cr3+, NO2-, Cr(NO2)3, aqueous phase, dissociate, molecular equation, water, cancelled, common ions, equation.
