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Let's analyze each chemical equation to determine if a precipitation reaction occurs by applying solubility rules. A precipitation reaction occurs when an insoluble product, or precipitate, is formed in the reaction.
Option A:
[tex]\[ \text{MgBr}_2 + 2 \text{HCl} \rightarrow \text{MgCl}_2 + 2 \text{HBr} \][/tex]
- MgBr[tex]\(_2\)[/tex]: Soluble (rule 3: bromides are soluble)
- HCl: Soluble (density of HCl in water, it's a strong acid)
- MgCl[tex]\(_2\)[/tex]: Soluble (rule 3: chlorides are soluble)
- HBr: Soluble (strong acid)
Since all reactants and products are soluble, no precipitate forms. Therefore, option A does not represent a precipitation reaction.
Option B:
[tex]\[ \text{K}_2\text{CO}_3 + \text{PbCl}_2 \rightarrow 2 \text{KCl} + \text{PbCO}_3 \][/tex]
- K[tex]\(_2\)[/tex]CO[tex]\(_3\)[/tex]: Soluble (rule 1: potassium compounds are soluble)
- PbCl[tex]\(_2\)[/tex]: Soluble with exceptions; slightly soluble in cold water, more soluble in hot water.
- KCl: Soluble (rule 1: potassium compounds are soluble)
- PbCO[tex]\(_3\)[/tex]: Insoluble (carbonates are generally insoluble except for those of alkali metals and ammonium)
Since PbCO[tex]\(_3\)[/tex] is insoluble, a precipitate forms. Therefore, option B represents a precipitation reaction.
Option C:
[tex]\[ 4 \text{LiC}_2\text{H}_3\text{O}_2 + \text{TiBr}_4 \rightarrow 4 \text{LiBr} + \text{Ti}\left(\text{C}_2\text{H}_3\text{O}_2\right)_4 \][/tex]
- LiC[tex]\(_2\)[/tex]H[tex]\(_3\)[/tex]O[tex]\(_2\)[/tex]: Soluble (rule 1: lithium compounds are soluble)
- TiBr[tex]\(_4\)[/tex]: Soluble (rule 3: bromides are soluble)
- LiBr: Soluble (rule 1: lithium compounds are soluble)
- Ti(C[tex]\(_2\)[/tex]H[tex]\(_3\)[/tex]O[tex]\(_2\)[/tex])[tex]\(_4\)[/tex]: Soluble (rule 2: acetates are soluble)
Since all reactants and products are soluble, no precipitate forms. Therefore, option C does not represent a precipitation reaction.
Option D:
[tex]\[ 2 \text{NH}_4\text{NO}_3 + \text{CuCl}_2 \rightarrow 2 \text{NH}_4\text{Cl} + \text{Cu(NO}_3)_2 \][/tex]
- NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex]: Soluble (rule 1: ammonium compounds are soluble)
- CuCl[tex]\(_2\)[/tex]: Soluble (rule 3: chlorides are soluble)
- NH[tex]\(_4\)[/tex]Cl: Soluble (rule 1: ammonium compounds are soluble)
- Cu(NO[tex]\(_3\)[/tex])[tex]\(_2\)[/tex]: Soluble (rule 2: nitrates are soluble)
Since all reactants and products are soluble, no precipitate forms. Therefore, option D does not represent a precipitation reaction.
Conclusion:
The chemical equation that represents a precipitation reaction is:
[tex]\[ \text{K}_2\text{CO}_3 + \text{PbCl}_2 \rightarrow 2 \text{KCl} + \text{PbCO}_3 \][/tex]
Hence, the correct answer is Option B.
Option A:
[tex]\[ \text{MgBr}_2 + 2 \text{HCl} \rightarrow \text{MgCl}_2 + 2 \text{HBr} \][/tex]
- MgBr[tex]\(_2\)[/tex]: Soluble (rule 3: bromides are soluble)
- HCl: Soluble (density of HCl in water, it's a strong acid)
- MgCl[tex]\(_2\)[/tex]: Soluble (rule 3: chlorides are soluble)
- HBr: Soluble (strong acid)
Since all reactants and products are soluble, no precipitate forms. Therefore, option A does not represent a precipitation reaction.
Option B:
[tex]\[ \text{K}_2\text{CO}_3 + \text{PbCl}_2 \rightarrow 2 \text{KCl} + \text{PbCO}_3 \][/tex]
- K[tex]\(_2\)[/tex]CO[tex]\(_3\)[/tex]: Soluble (rule 1: potassium compounds are soluble)
- PbCl[tex]\(_2\)[/tex]: Soluble with exceptions; slightly soluble in cold water, more soluble in hot water.
- KCl: Soluble (rule 1: potassium compounds are soluble)
- PbCO[tex]\(_3\)[/tex]: Insoluble (carbonates are generally insoluble except for those of alkali metals and ammonium)
Since PbCO[tex]\(_3\)[/tex] is insoluble, a precipitate forms. Therefore, option B represents a precipitation reaction.
Option C:
[tex]\[ 4 \text{LiC}_2\text{H}_3\text{O}_2 + \text{TiBr}_4 \rightarrow 4 \text{LiBr} + \text{Ti}\left(\text{C}_2\text{H}_3\text{O}_2\right)_4 \][/tex]
- LiC[tex]\(_2\)[/tex]H[tex]\(_3\)[/tex]O[tex]\(_2\)[/tex]: Soluble (rule 1: lithium compounds are soluble)
- TiBr[tex]\(_4\)[/tex]: Soluble (rule 3: bromides are soluble)
- LiBr: Soluble (rule 1: lithium compounds are soluble)
- Ti(C[tex]\(_2\)[/tex]H[tex]\(_3\)[/tex]O[tex]\(_2\)[/tex])[tex]\(_4\)[/tex]: Soluble (rule 2: acetates are soluble)
Since all reactants and products are soluble, no precipitate forms. Therefore, option C does not represent a precipitation reaction.
Option D:
[tex]\[ 2 \text{NH}_4\text{NO}_3 + \text{CuCl}_2 \rightarrow 2 \text{NH}_4\text{Cl} + \text{Cu(NO}_3)_2 \][/tex]
- NH[tex]\(_4\)[/tex]NO[tex]\(_3\)[/tex]: Soluble (rule 1: ammonium compounds are soluble)
- CuCl[tex]\(_2\)[/tex]: Soluble (rule 3: chlorides are soluble)
- NH[tex]\(_4\)[/tex]Cl: Soluble (rule 1: ammonium compounds are soluble)
- Cu(NO[tex]\(_3\)[/tex])[tex]\(_2\)[/tex]: Soluble (rule 2: nitrates are soluble)
Since all reactants and products are soluble, no precipitate forms. Therefore, option D does not represent a precipitation reaction.
Conclusion:
The chemical equation that represents a precipitation reaction is:
[tex]\[ \text{K}_2\text{CO}_3 + \text{PbCl}_2 \rightarrow 2 \text{KCl} + \text{PbCO}_3 \][/tex]
Hence, the correct answer is Option B.
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