Respuesta :
Answer:
Here's what I get
Explanation:
Solubility rules
- Salts containing halides are generally soluble. Important exceptions to this rule are halides of silver, mercury, and lead(II).
- All acetates, chlorates, and perchlorates are soluble
So, PbCl₂ is insoluble, and Pb(ClO₃)₂ is soluble.
1. "Molecular" equation
[tex]\rm Pb(ClO_{3})_{2}(aq) + 2HCl(aq) \longrightarrow \, PbCl_{2}(s) + 2HClO_{3}(aq)[/tex]
2. Ionic equation
Convert the soluble salts to their hydrated ions.
HCl and HClO₃ are strong acids. Convert them to their ions.
[tex]\rm Pb^{2+}(aq) + 2ClO_{3}^{-}(aq)+ 2H^{+}(aq) + 2Cl^{-}(aq) \longrightarrow \, PbCl_{2}(s) + 2H^{+}(aq) + 2ClO_{3}^{-}(aq)[/tex]
3. Net ionic equation
Cancel all ions that appear on both sides of the reaction arrow (in boldface).
[tex]\rm {Pb}^{2+}(aq) + \textbf{2ClO}_{3}^{-}(aq)+ \textbf{2H}^{+}(aq) + 2Cl^{-}(aq) \longrightarrow \, PbCl_{2}(s) + \textbf{2H}^{+}(aq) + \textbf{2ClO}_{3}^{-}(aq)[/tex]
The net ionic equation is
[tex]\rm {Pb}^{2+}(aq) + 2Cl^{-}(aq) \longrightarrow \, PbCl_{2}(s)[/tex]
4. Theoretical yield
We have the volumes and concentrations of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
(i). Gather all the information in one place with molar masses above the formulas and masses below them.
M_r: 278.11
Pb(ClO₃)₂ + 2HCl ⟶ PbCl₂ + 2HClO₃
Volume/mL: 125 95
c/mol·L⁻¹: 0.85 0.85
(ii) Calculate the moles of each reactant
[tex]\text{Moles of Pb(ClO$_{3}$)}_{2} = \text{0.125 L} \times \dfrac{\text{0.85 mol}}{\text{1 L }} = \text{0.1062 mol}\\\text{Moles of HCl} = \text{0.095 L} \times \dfrac{\text{0.85 mol}}{\text{1 L }} = \text{0.08075 mol}[/tex]
(iii) Identify the limiting reactant
Calculate the moles of PbCl₂ we can obtain from each reactant.
From Pb(ClO₃)₂:
The molar ratio of PbCl₂:Pb(ClO₃)₂ is 2:2
Moles of PbCl₂ = 0.1062 × 2/2 =0.1062 mol PbCl₂
From HCl :
The molar ratio of PbCl₂:HCl is 1 mol PbCl₂:2 mol HCl.
Moles of PbCl₂ = 0.08075 × 1/2 = 0.04038 mol PbCl₂
The limiting reactant is HCl because it gives the smaller amount of PbCl₂.
(iv) Calculate the theoretical yield of PbCl₂.
[tex]\text{Theor. yield of PbCl}_{2} = \text{0.0438 mol} \times \dfrac{\text{278.11 g}}{\text{ 1 mol}} = \textbf{11.2 g}[/tex]
5. Calculate the actual yield of PbCl₂
[tex]\text{Actual yield} = \text{11.2 g theor.} \times \dfrac{\text{ 68 g actual}}{\text{100 g theor,}} = \textbf{7.6 g}[/tex]
6. Calculate [ClO₃⁻]
Original concentration of Pb(ClO₃)₂ = 0.85 mol·L⁻¹
Original concentration of ClO₃ = 2 × 0.85 = 1.70 mol·L⁻¹
The solution was diluted by the addition of HCl.
Total volume = 125 + 95 =220 mL
c₁V₁ = c₂V₂
1.70 mol·L⁻¹ × 125 mL = c₂ × 220 mL
212.5 mol·L⁻¹ = 200 c₂
c₂ = (212.5 mL)/200 = 1.06 mol·L⁻¹
7. Calculate [Pb²⁺].
Moles of Pb²⁺ originally present = 0.1062 mol
Moles of Pb²⁺removed = 0.04038 mol
Moles of Pb²⁺ remaining = 0.0659 mol
c = 0.0659 mol/0.220 L = 0.299 mol·L⁻¹
