Question

pb(oh)₂ reacts with hcl in a double - displacement reaction, what is the balanced equation?
○ pb(oh)₂ + 2hcl → h₂pb + 2cl(oh)
○ 2h₂o + pbcl₂ → pb(oh)₂ + 2hcl
○ pb(oh)₂ + hcl → h₂o + pbcl
○ pb(oh)₂ + 2hcl → 2h₂o + pbcl₂

Explanation:

Step1: Understand double - displacement reaction

In a double - displacement reaction, the cations and anions of two compounds exchange partners. For \(Pb(OH)_2\) (a base) and \(HCl\) (an acid), the general form is \(AB + CD
ightarrow AD+CB\), where \(A = Pb^{2 +}\), \(B=(OH)^{-}\), \(C = H^{+}\), \(D = Cl^{-}\). The products should be \(H_2O\) (from \(H^{+}\) and \(OH^{-}\)) and \(PbCl_2\) (from \(Pb^{2+}\) and \(Cl^{-}\)).

Step2: Balance the equation

• For \(Pb\): There is 1 \(Pb\) atom on both sides initially.
• For \(Cl\): In \(HCl\), to get \(PbCl_2\) (which has 2 \(Cl\) atoms), we need 2 \(HCl\) molecules.
• For \(H\) and \(O\):
• Left - hand side: In \(Pb(OH)_2\) there are 2 \(O\) and 2 \(H\) (from \(OH\) groups) and 2 \(H\) from \(2HCl\), total \(H = 4\) and \(O=2\).
• Right - hand side: In \(2H_2O\) there are 4 \(H\) and 2 \(O\).

The balanced equation is \(Pb(OH)_2+2HCl
ightarrow 2H_2O + PbCl_2\)

Answer:

\(Pb(OH)_2+2HCl
ightarrow 2H_2O + PbCl_2\) (the fourth option)