Answer :
To determine which chemical equation represents a precipitation reaction, we need to identify the formation of an insoluble solid (precipitate) as a product. A precipitation reaction involves the mixing of two aqueous solutions resulting in a solid precipitate.
Let's evaluate each option:
A. [tex]\( MgBr_2 + 2 HCl \rightarrow MgCl_2 + 2 HBr \)[/tex]
- Here, both [tex]\( MgBr_2 \)[/tex] and [tex]\( HCl \)[/tex] are soluble in water, and the products, [tex]\( MgCl_2 \)[/tex] and [tex]\( HBr \)[/tex], are also soluble in water. No precipitate is formed in this reaction.
B. [tex]\( K_2CO_3 + PbCl_2 \rightarrow 2 KCl + PbCO_3 \)[/tex]
- In this reaction, [tex]\( K_2CO_3 \)[/tex] and [tex]\( PbCl_2 \)[/tex] are soluble in water. However, [tex]\( PbCO_3 \)[/tex] (Lead(II) carbonate) is an insoluble compound and forms a solid precipitate. Thus, this reaction forms a precipitate.
C. [tex]\( 4 LiC_2H_3O_2 + TiBr_4 \rightarrow 4 LiBr + Ti(C_2H_3O_2)_4 \)[/tex]
- Both [tex]\( LiC_2H_3O_2 \)[/tex] (lithium acetate) and [tex]\( TiBr_4 \)[/tex] (titanium(IV) bromide) are soluble, and the products [tex]\( LiBr \)[/tex] and [tex]\( Ti(C_2H_3O_2)_4 \)[/tex] are also soluble. No precipitate is formed.
D. [tex]\( 2 NH_4NO_3 + CuCl_2 \rightarrow 2 NH_4Cl + Cu(NO_3)_2 \)[/tex]
- Both [tex]\( NH_4NO_3 \)[/tex] (ammonium nitrate) and [tex]\( CuCl_2 \)[/tex] (copper(II) chloride) are soluble in water, and so are the products [tex]\( NH_4Cl \)[/tex] and [tex]\( Cu(NO_3)_2 \)[/tex]. No precipitate is formed.
Therefore, the correct chemical equation representing a precipitation reaction is:
B. [tex]\( K_2CO_3 + PbCl_2 \rightarrow 2 KCl + PbCO_3 \)[/tex]
Let's evaluate each option:
A. [tex]\( MgBr_2 + 2 HCl \rightarrow MgCl_2 + 2 HBr \)[/tex]
- Here, both [tex]\( MgBr_2 \)[/tex] and [tex]\( HCl \)[/tex] are soluble in water, and the products, [tex]\( MgCl_2 \)[/tex] and [tex]\( HBr \)[/tex], are also soluble in water. No precipitate is formed in this reaction.
B. [tex]\( K_2CO_3 + PbCl_2 \rightarrow 2 KCl + PbCO_3 \)[/tex]
- In this reaction, [tex]\( K_2CO_3 \)[/tex] and [tex]\( PbCl_2 \)[/tex] are soluble in water. However, [tex]\( PbCO_3 \)[/tex] (Lead(II) carbonate) is an insoluble compound and forms a solid precipitate. Thus, this reaction forms a precipitate.
C. [tex]\( 4 LiC_2H_3O_2 + TiBr_4 \rightarrow 4 LiBr + Ti(C_2H_3O_2)_4 \)[/tex]
- Both [tex]\( LiC_2H_3O_2 \)[/tex] (lithium acetate) and [tex]\( TiBr_4 \)[/tex] (titanium(IV) bromide) are soluble, and the products [tex]\( LiBr \)[/tex] and [tex]\( Ti(C_2H_3O_2)_4 \)[/tex] are also soluble. No precipitate is formed.
D. [tex]\( 2 NH_4NO_3 + CuCl_2 \rightarrow 2 NH_4Cl + Cu(NO_3)_2 \)[/tex]
- Both [tex]\( NH_4NO_3 \)[/tex] (ammonium nitrate) and [tex]\( CuCl_2 \)[/tex] (copper(II) chloride) are soluble in water, and so are the products [tex]\( NH_4Cl \)[/tex] and [tex]\( Cu(NO_3)_2 \)[/tex]. No precipitate is formed.
Therefore, the correct chemical equation representing a precipitation reaction is:
B. [tex]\( K_2CO_3 + PbCl_2 \rightarrow 2 KCl + PbCO_3 \)[/tex]
