To determine which chemical equation represents a precipitation reaction, we must identify the reaction where an insoluble solid (precipitate) forms. Let's analyze each option:
Option A:
[tex]\[ \text{MgBr}_2 + 2 \text{HCl} \rightarrow \text{MgCl}_2 + 2 \text{HBr} \][/tex]
In this reaction, all products (MgCl₂ and HBr) are soluble in water, so no precipitate forms.
Option B:
[tex]\[ \text{K}_2\text{CO}_3 + \text{PbCl}_2 \rightarrow 2 \text{KCl} + \text{PbCO}_3 \][/tex]
In this reaction, potassium chloride (KCl) is soluble in water, but lead(II) carbonate (PbCO₃) is not soluble in water and thus forms a precipitate.
Option C:
[tex]\[ 4 \text{LiC}_2\text{H}_3\text{O}_2 + \text{TiBr}_4 \rightarrow 4 \text{LiBr} + \text{Ti}( \text{C}_2\text{H}_3\text{O}_2)_4 \][/tex]
All products (LiBr and the complex Ti(C₂H₃O₂)₄) are soluble in water, so no precipitate forms.
Option D:
[tex]\[ 2 \text{NH}_4\text{NO}_3 + \text{CuCl}_2 \rightarrow 2 \text{NH}_4\text{Cl} + \text{Cu}( \text{NO}_3)_2 \][/tex]
All products (NH₄Cl and Cu(NO₃)₂) are soluble in water, so no precipitate forms.
After examining all the options, only Option B:
[tex]\[ \text{K}_2\text{CO}_3 + \text{PbCl}_2 \rightarrow 2 \text{KCl} + \text{PbCO}_3 \][/tex]
results in the formation of an insoluble solid, lead(II) carbonate (PbCO₃), which precipitates out of the solution.
Therefore, the correct answer is:
B. [tex]\( \text{K}_2\text{CO}_3 + \text{PbCl}_2 \rightarrow 2 \text{KCl} + \text{PbCO}_3 \)[/tex]
