Question

question 39 (1 point) listen in the reaction of aqueous lead(ii)nitrate and aqueous ammonium chloride, the net ionic equation for the precipitate formed will be: 1) fe²⁺(aq) + 2cl⁻(aq)→ fecl₂(aq) 2) pb²⁺(aq) + no₃⁻(aq) → pb(no₃)₂(aq) 3) pb²⁺ (aq) + 2cl⁻(aq) → pbcl₂(s) 4) nh₄⁺(aq) + no₃⁻(aq) → nh₄no₃(aq)

Explanation:

Step1: Identify Reactants

The reactants are aqueous Lead(II) nitrate ($\ce{Pb(NO3)2(aq)}$) and aqueous ammonium chloride ($\ce{NH4Cl(aq)}$). Dissociating them: $\ce{Pb(NO3)2(aq) -> Pb^{2+}(aq) + 2NO3^-(aq)}$ and $\ce{NH4Cl(aq) -> NH4^+(aq) + Cl^-(aq)}$.

Step2: Determine Products

Possible products: $\ce{PbCl2}$ and $\ce{NH4NO3}$. $\ce{NH4NO3}$ is soluble (aqueous), $\ce{PbCl2}$ is insoluble (precipitate, solid).

Step3: Write Net Ionic Equation

Spectator ions ($\ce{NO3^-}$, $\ce{NH4^+}$) are removed. So, $\ce{Pb^{2+}(aq) + 2Cl^-(aq) -> PbCl2(s)}$.

Step4: Evaluate Options

• Option 1: Involves $\ce{Fe^{2+}}$, not in reactants. Incorrect.
• Option 2: $\ce{Pb(NO3)2}$ is soluble, no precipitate. Incorrect.
• Option 3: Matches net ionic equation for $\ce{PbCl2}$ precipitate. Correct.
• Option 4: $\ce{NH4NO3}$ is soluble, no precipitate. Incorrect.

Answer:

1. $\ce{Pb^{2+}(aq) + 2Cl^-(aq) -> PbCl2(s)}$