Answer :
To determine which equations represent precipitation reactions, we need to check for the formation of insoluble products based on the solubility rules provided. Solubility rules will help us identify which products will precipitate (form an insoluble solid in a solution). Let’s analyze each equation step-by-step:
1. [tex]\( Na_2S + FeBr_2 \rightarrow 2 NaBr + FeS \)[/tex]
- [tex]\( Na_2S \)[/tex]: Sodium sulfide, soluble (rule 1)
- [tex]\( FeBr_2 \)[/tex]: Iron(II) bromide, soluble (rule 3 does not mention iron)
- [tex]\( 2 NaBr \)[/tex]: Sodium bromide, soluble (rule 1)
- [tex]\( FeS \)[/tex]: Iron(II) sulfide, generally insoluble (sulfides are usually insoluble except with group 1 metals and ammonium)
This reaction forms FeS, an insoluble solid, which makes it a precipitation reaction.
2. [tex]\( MgSO_4 + CaCl_2 \rightarrow MgCl_2 + CaSO_4 \)[/tex]
- [tex]\( MgSO_4 \)[/tex]: Magnesium sulfate, soluble (rule 4 does not mention magnesium)
- [tex]\( CaCl_2 \)[/tex]: Calcium chloride, soluble (rule 3)
- [tex]\( MgCl_2 \)[/tex]: Magnesium chloride, soluble (rule 3 does not mention magnesium)
- [tex]\( CaSO_4 \)[/tex]: Calcium sulfate, sparingly soluble – tends to form a precipitate because sulfates with calcium are exceptions (rule 4)
This reaction forms CaSO_4, which is sparingly soluble and thus forms a precipitate, making it a precipitation reaction.
3. [tex]\( LiOH + NH_4I \rightarrow LiI + NH_4OH \)[/tex]
- [tex]\( LiOH \)[/tex]: Lithium hydroxide, soluble (rule 1)
- [tex]\( NH_4I \)[/tex]: Ammonium iodide, soluble (rule 1)
- [tex]\( LiI \)[/tex]: Lithium iodide, soluble (rule 1 and rule 3)
- [tex]\( NH_4OH \)[/tex]: Ammonium hydroxide, soluble (rule 1)
All products are soluble. This does not form a precipitate; hence, it is not a precipitation reaction.
4. [tex]\( 2 NaCl + K_2S \rightarrow Na_2S + 2 KCl \)[/tex]
- [tex]\( 2 NaCl \)[/tex]: Sodium chloride, soluble (rule 1)
- [tex]\( K_2S \)[/tex]: Potassium sulfide, soluble (rule 1)
- [tex]\( Na_2S \)[/tex]: Sodium sulfide, soluble (rule 1)
- [tex]\( 2 KCl \)[/tex]: Potassium chloride, soluble (rule 1)
All products are soluble. This does not form a precipitate; hence, it is not a precipitation reaction.
5. [tex]\( AgNO_3 + NaCl \rightarrow AgCl + NaNO_3 \)[/tex]
- [tex]\( AgNO_3 \)[/tex]: Silver nitrate, soluble (rule 2)
- [tex]\( NaCl \)[/tex]: Sodium chloride, soluble (rule 1)
- [tex]\( AgCl \)[/tex]: Silver chloride, insoluble (rule 3, exception to chlorides)
- [tex]\( NaNO_3 \)[/tex]: Sodium nitrate, soluble (rule 1 and rule 2)
This reaction forms AgCl, an insoluble solid, which makes it a precipitation reaction.
Summary of precipitation reactions:
- [tex]\( Na_2 S + FeBr_2 \rightarrow 2 NaBr + FeS \)[/tex]
- [tex]\( MgSO_4 + CaCl_2 \rightarrow MgCl_2 + CaSO_4 \)[/tex]
- [tex]\( AgNO_3 + NaCl \rightarrow AgCl + NaNO_3 \)[/tex]
Hence, the correct equations representing precipitation reactions are:
- [tex]\( Na_2 S + FeBr_2 \rightarrow 2 NaBr + FeS \)[/tex]
- [tex]\( MgSO_4 + CaCl_2 \rightarrow MgCl_2 + CaSO_4 \)[/tex]
- [tex]\( AgNO_3 + NaCl \rightarrow AgCl + NaNO_3 \)[/tex]
1. [tex]\( Na_2S + FeBr_2 \rightarrow 2 NaBr + FeS \)[/tex]
- [tex]\( Na_2S \)[/tex]: Sodium sulfide, soluble (rule 1)
- [tex]\( FeBr_2 \)[/tex]: Iron(II) bromide, soluble (rule 3 does not mention iron)
- [tex]\( 2 NaBr \)[/tex]: Sodium bromide, soluble (rule 1)
- [tex]\( FeS \)[/tex]: Iron(II) sulfide, generally insoluble (sulfides are usually insoluble except with group 1 metals and ammonium)
This reaction forms FeS, an insoluble solid, which makes it a precipitation reaction.
2. [tex]\( MgSO_4 + CaCl_2 \rightarrow MgCl_2 + CaSO_4 \)[/tex]
- [tex]\( MgSO_4 \)[/tex]: Magnesium sulfate, soluble (rule 4 does not mention magnesium)
- [tex]\( CaCl_2 \)[/tex]: Calcium chloride, soluble (rule 3)
- [tex]\( MgCl_2 \)[/tex]: Magnesium chloride, soluble (rule 3 does not mention magnesium)
- [tex]\( CaSO_4 \)[/tex]: Calcium sulfate, sparingly soluble – tends to form a precipitate because sulfates with calcium are exceptions (rule 4)
This reaction forms CaSO_4, which is sparingly soluble and thus forms a precipitate, making it a precipitation reaction.
3. [tex]\( LiOH + NH_4I \rightarrow LiI + NH_4OH \)[/tex]
- [tex]\( LiOH \)[/tex]: Lithium hydroxide, soluble (rule 1)
- [tex]\( NH_4I \)[/tex]: Ammonium iodide, soluble (rule 1)
- [tex]\( LiI \)[/tex]: Lithium iodide, soluble (rule 1 and rule 3)
- [tex]\( NH_4OH \)[/tex]: Ammonium hydroxide, soluble (rule 1)
All products are soluble. This does not form a precipitate; hence, it is not a precipitation reaction.
4. [tex]\( 2 NaCl + K_2S \rightarrow Na_2S + 2 KCl \)[/tex]
- [tex]\( 2 NaCl \)[/tex]: Sodium chloride, soluble (rule 1)
- [tex]\( K_2S \)[/tex]: Potassium sulfide, soluble (rule 1)
- [tex]\( Na_2S \)[/tex]: Sodium sulfide, soluble (rule 1)
- [tex]\( 2 KCl \)[/tex]: Potassium chloride, soluble (rule 1)
All products are soluble. This does not form a precipitate; hence, it is not a precipitation reaction.
5. [tex]\( AgNO_3 + NaCl \rightarrow AgCl + NaNO_3 \)[/tex]
- [tex]\( AgNO_3 \)[/tex]: Silver nitrate, soluble (rule 2)
- [tex]\( NaCl \)[/tex]: Sodium chloride, soluble (rule 1)
- [tex]\( AgCl \)[/tex]: Silver chloride, insoluble (rule 3, exception to chlorides)
- [tex]\( NaNO_3 \)[/tex]: Sodium nitrate, soluble (rule 1 and rule 2)
This reaction forms AgCl, an insoluble solid, which makes it a precipitation reaction.
Summary of precipitation reactions:
- [tex]\( Na_2 S + FeBr_2 \rightarrow 2 NaBr + FeS \)[/tex]
- [tex]\( MgSO_4 + CaCl_2 \rightarrow MgCl_2 + CaSO_4 \)[/tex]
- [tex]\( AgNO_3 + NaCl \rightarrow AgCl + NaNO_3 \)[/tex]
Hence, the correct equations representing precipitation reactions are:
- [tex]\( Na_2 S + FeBr_2 \rightarrow 2 NaBr + FeS \)[/tex]
- [tex]\( MgSO_4 + CaCl_2 \rightarrow MgCl_2 + CaSO_4 \)[/tex]
- [tex]\( AgNO_3 + NaCl \rightarrow AgCl + NaNO_3 \)[/tex]