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When aqueous solutions of sodium sulfide (Na[tex]\(_2\)[/tex]S) and iron(II) nitrate (Fe(NO[tex]\(_3\)[/tex])[tex]\(_2\)[/tex]) are mixed, a reaction occurs, resulting in the formation of a precipitate of iron(II) sulfide (FeS).
The complete ionic equation for this reaction can be derived as follows:
1. Identify the reactants and their dissociation in water:
- Sodium sulfide (Na[tex]\(_2\)[/tex]S) dissociates into sodium ions (Na[tex]\(^+\)[/tex]) and sulfide ions (S[tex]\(^{2-}\)[/tex]):
[tex]\[ \text{Na}_2\text{S} \rightarrow 2 \text{Na}^+(aq) + \text{S}^{2-}(aq) \][/tex]
- Iron(II) nitrate (Fe(NO[tex]\(_3\)[/tex])[tex]\(_2\)[/tex]) dissociates into iron(II) ions (Fe[tex]\(^{2+}\)[/tex]) and nitrate ions (NO[tex]\(_3^{-}\)[/tex]):
[tex]\[ \text{Fe(NO}_3)_2 \rightarrow \text{Fe}^{2+}(aq) + 2 \text{NO}_3^{-}(aq) \][/tex]
2. Write the ions of the reactants in aqueous solution:
[tex]\[ 2 \text{Na}^+(aq) + \text{S}^{2-}(aq) + \text{Fe}^{2+}(aq) + 2 \text{NO}_3^{-}(aq) \][/tex]
3. Determine the products of the reaction:
- When these ions interact, iron(II) ions (Fe[tex]\(^{2+}\)[/tex]) react with sulfide ions (S[tex]\(^{2-}\)[/tex]) to form iron(II) sulfide (FeS), which is insoluble in water and precipitates out.
- Sodium ions (Na[tex]\(^+\)[/tex]) and nitrate ions (NO[tex]\(_3^{-}\)[/tex]) remain in solution as spectator ions.
4. Write the complete ionic equation:
[tex]\[ 2 \text{Na}^+(aq) + \text{S}^{2-}(aq) + \text{Fe}^{2+}(aq) + 2 \text{NO}_3^{-}(aq) \rightarrow \text{FeS}(s) + 2 \text{Na}^+(aq) + 2 \text{NO}_3^{-}(aq) \][/tex]
This equation shows the formation of the precipitate FeS, while Na[tex]\(^+\)[/tex] and NO[tex]\(_3^{-}\)[/tex] remain in the solution. Therefore, the correct answer is:
[tex]\[ 2 \text{Na}^+(aq) + \text{S}^{2-}(aq) + \text{Fe}^{2+}(aq) + 2 \text{NO}_3^{-}(aq) \rightarrow \text{FeS}(s) + 2 \text{Na}^+(aq) + 2 \text{NO}_3^{-}(aq) \][/tex]
This is the complete ionic equation for the reaction that occurs when aqueous solutions of sodium sulfide and iron(II) nitrate are mixed.
The complete ionic equation for this reaction can be derived as follows:
1. Identify the reactants and their dissociation in water:
- Sodium sulfide (Na[tex]\(_2\)[/tex]S) dissociates into sodium ions (Na[tex]\(^+\)[/tex]) and sulfide ions (S[tex]\(^{2-}\)[/tex]):
[tex]\[ \text{Na}_2\text{S} \rightarrow 2 \text{Na}^+(aq) + \text{S}^{2-}(aq) \][/tex]
- Iron(II) nitrate (Fe(NO[tex]\(_3\)[/tex])[tex]\(_2\)[/tex]) dissociates into iron(II) ions (Fe[tex]\(^{2+}\)[/tex]) and nitrate ions (NO[tex]\(_3^{-}\)[/tex]):
[tex]\[ \text{Fe(NO}_3)_2 \rightarrow \text{Fe}^{2+}(aq) + 2 \text{NO}_3^{-}(aq) \][/tex]
2. Write the ions of the reactants in aqueous solution:
[tex]\[ 2 \text{Na}^+(aq) + \text{S}^{2-}(aq) + \text{Fe}^{2+}(aq) + 2 \text{NO}_3^{-}(aq) \][/tex]
3. Determine the products of the reaction:
- When these ions interact, iron(II) ions (Fe[tex]\(^{2+}\)[/tex]) react with sulfide ions (S[tex]\(^{2-}\)[/tex]) to form iron(II) sulfide (FeS), which is insoluble in water and precipitates out.
- Sodium ions (Na[tex]\(^+\)[/tex]) and nitrate ions (NO[tex]\(_3^{-}\)[/tex]) remain in solution as spectator ions.
4. Write the complete ionic equation:
[tex]\[ 2 \text{Na}^+(aq) + \text{S}^{2-}(aq) + \text{Fe}^{2+}(aq) + 2 \text{NO}_3^{-}(aq) \rightarrow \text{FeS}(s) + 2 \text{Na}^+(aq) + 2 \text{NO}_3^{-}(aq) \][/tex]
This equation shows the formation of the precipitate FeS, while Na[tex]\(^+\)[/tex] and NO[tex]\(_3^{-}\)[/tex] remain in the solution. Therefore, the correct answer is:
[tex]\[ 2 \text{Na}^+(aq) + \text{S}^{2-}(aq) + \text{Fe}^{2+}(aq) + 2 \text{NO}_3^{-}(aq) \rightarrow \text{FeS}(s) + 2 \text{Na}^+(aq) + 2 \text{NO}_3^{-}(aq) \][/tex]
This is the complete ionic equation for the reaction that occurs when aqueous solutions of sodium sulfide and iron(II) nitrate are mixed.
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