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.
