Answer :
Certainly! Let's analyze each of the given chemical equations to determine which ones represent precipitation reactions:
1. [tex]\( \text{Na}_2\text{S} + \text{FeBr}_2 \rightarrow 2 \, \text{NaBr} + \text{FeS} \)[/tex]
- In this reaction, sodium sulfide ([tex]\(\text{Na}_2\text{S}\)[/tex]) reacts with iron(II) bromide ([tex]\(\text{FeBr}_2\)[/tex]) to form sodium bromide ([tex]\(\text{NaBr}\)[/tex]) and iron(II) sulfide ([tex]\(\text{FeS}\)[/tex]). According to solubility rules, [tex]\(\text{FeS}\)[/tex] is an insoluble solid, meaning that it forms a precipitate. Therefore, this is a precipitation reaction.
2. [tex]\( \text{MgSO}_4 + \text{CaCl}_2 \rightarrow \text{MgCl}_2 + \text{CaSO}_4 \)[/tex]
- In this reaction, magnesium sulfate ([tex]\(\text{MgSO}_4\)[/tex]) reacts with calcium chloride ([tex]\(\text{CaCl}_2\)[/tex]) to form magnesium chloride ([tex]\(\text{MgCl}_2\)[/tex]) and calcium sulfate ([tex]\(\text{CaSO}_4\)[/tex]). According to solubility rules, [tex]\(\text{CaSO}_4\)[/tex] is an insoluble solid, meaning that it forms a precipitate. Therefore, this is a precipitation reaction.
3. [tex]\( \text{LiOH} + \text{NH}_4\text{I} \rightarrow \text{LiI} + \text{NH}_4\text{OH} \)[/tex]
- In this reaction, lithium hydroxide ([tex]\(\text{LiOH}\)[/tex]) reacts with ammonium iodide ([tex]\(\text{NH}_4\text{I}\)[/tex]) to form lithium iodide ([tex]\(\text{LiI}\)[/tex]) and ammonium hydroxide ([tex]\(\text{NH}_4\text{OH}\)[/tex]). Ammonium hydroxide ([tex]\(\text{NH}_4\text{OH}\)[/tex]) is a weak base and remains in solution, not forming a precipitate. Therefore, this is not a precipitation reaction.
4. [tex]\( 2 \, \text{NaCl} + \text{K}_2\text{S} \rightarrow \text{Na}_2\text{S} + 2 \, \text{KCl} \)[/tex]
- In this reaction, sodium chloride ([tex]\(\text{NaCl}\)[/tex]) reacts with potassium sulfide ([tex]\(\text{K}_2\text{S}\)[/tex]) to form sodium sulfide ([tex]\(\text{Na}_2\text{S}\)[/tex]) and potassium chloride ([tex]\(\text{KCl}\)[/tex]). All reactants and products are soluble in water, meaning no precipitate is formed. Therefore, this is not a precipitation reaction.
5. [tex]\( \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3 \)[/tex]
- In this reaction, silver nitrate ([tex]\(\text{AgNO}_3\)[/tex]) reacts with sodium chloride ([tex]\(\text{NaCl}\)[/tex]) to form silver chloride ([tex]\(\text{AgCl}\)[/tex]) and sodium nitrate ([tex]\(\text{NaNO}_3\)[/tex]). According to solubility rules, [tex]\(\text{AgCl}\)[/tex] is an insoluble solid, meaning that it forms a precipitate. Therefore, this is a precipitation reaction.
Based on the above analyses, the correct equations that represent precipitation reactions are:
1. [tex]\( \text{Na}_2\text{S} + \text{FeBr}_2 \rightarrow 2 \, \text{NaBr} + \text{FeS} \)[/tex]
2. [tex]\( \text{MgSO}_4 + \text{CaCl}_2 \rightarrow \text{MgCl}_2 + \text{CaSO}_4 \)[/tex]
5. [tex]\( \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3 \)[/tex]
So the equations representing precipitation reactions are:
- [1, 2, 5]
1. [tex]\( \text{Na}_2\text{S} + \text{FeBr}_2 \rightarrow 2 \, \text{NaBr} + \text{FeS} \)[/tex]
- In this reaction, sodium sulfide ([tex]\(\text{Na}_2\text{S}\)[/tex]) reacts with iron(II) bromide ([tex]\(\text{FeBr}_2\)[/tex]) to form sodium bromide ([tex]\(\text{NaBr}\)[/tex]) and iron(II) sulfide ([tex]\(\text{FeS}\)[/tex]). According to solubility rules, [tex]\(\text{FeS}\)[/tex] is an insoluble solid, meaning that it forms a precipitate. Therefore, this is a precipitation reaction.
2. [tex]\( \text{MgSO}_4 + \text{CaCl}_2 \rightarrow \text{MgCl}_2 + \text{CaSO}_4 \)[/tex]
- In this reaction, magnesium sulfate ([tex]\(\text{MgSO}_4\)[/tex]) reacts with calcium chloride ([tex]\(\text{CaCl}_2\)[/tex]) to form magnesium chloride ([tex]\(\text{MgCl}_2\)[/tex]) and calcium sulfate ([tex]\(\text{CaSO}_4\)[/tex]). According to solubility rules, [tex]\(\text{CaSO}_4\)[/tex] is an insoluble solid, meaning that it forms a precipitate. Therefore, this is a precipitation reaction.
3. [tex]\( \text{LiOH} + \text{NH}_4\text{I} \rightarrow \text{LiI} + \text{NH}_4\text{OH} \)[/tex]
- In this reaction, lithium hydroxide ([tex]\(\text{LiOH}\)[/tex]) reacts with ammonium iodide ([tex]\(\text{NH}_4\text{I}\)[/tex]) to form lithium iodide ([tex]\(\text{LiI}\)[/tex]) and ammonium hydroxide ([tex]\(\text{NH}_4\text{OH}\)[/tex]). Ammonium hydroxide ([tex]\(\text{NH}_4\text{OH}\)[/tex]) is a weak base and remains in solution, not forming a precipitate. Therefore, this is not a precipitation reaction.
4. [tex]\( 2 \, \text{NaCl} + \text{K}_2\text{S} \rightarrow \text{Na}_2\text{S} + 2 \, \text{KCl} \)[/tex]
- In this reaction, sodium chloride ([tex]\(\text{NaCl}\)[/tex]) reacts with potassium sulfide ([tex]\(\text{K}_2\text{S}\)[/tex]) to form sodium sulfide ([tex]\(\text{Na}_2\text{S}\)[/tex]) and potassium chloride ([tex]\(\text{KCl}\)[/tex]). All reactants and products are soluble in water, meaning no precipitate is formed. Therefore, this is not a precipitation reaction.
5. [tex]\( \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3 \)[/tex]
- In this reaction, silver nitrate ([tex]\(\text{AgNO}_3\)[/tex]) reacts with sodium chloride ([tex]\(\text{NaCl}\)[/tex]) to form silver chloride ([tex]\(\text{AgCl}\)[/tex]) and sodium nitrate ([tex]\(\text{NaNO}_3\)[/tex]). According to solubility rules, [tex]\(\text{AgCl}\)[/tex] is an insoluble solid, meaning that it forms a precipitate. Therefore, this is a precipitation reaction.
Based on the above analyses, the correct equations that represent precipitation reactions are:
1. [tex]\( \text{Na}_2\text{S} + \text{FeBr}_2 \rightarrow 2 \, \text{NaBr} + \text{FeS} \)[/tex]
2. [tex]\( \text{MgSO}_4 + \text{CaCl}_2 \rightarrow \text{MgCl}_2 + \text{CaSO}_4 \)[/tex]
5. [tex]\( \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3 \)[/tex]
So the equations representing precipitation reactions are:
- [1, 2, 5]
