Answer:
[tex]\large \boxed{84.7 \, \%}[/tex]
Explanation:
Mᵣ: 58.44 278.11
Pb(NO₃)₂ + 2NaCl ⟶ PbCl₂ + 2NaNO₃
m/g: 26.3
1. Moles of NaCl
[tex]\text{Moles of NaCl} = \text{26.3 g NaCl} \times \dfrac{\text{1 mol NaCl}}{\text{58.44 g NaCl}} = \text{0.4505 mol NaCl}[/tex]
(b) Moles of PbCl₂
[tex]\text{Moles of PbCl${_2}$} = \text{0.4505 mol NaCl} \times \dfrac{\text{1 mol PbCl${_2}$}}{\text{2 mol NaCl}} = \text{0.2253 mol PbCl${_2}$}[/tex]
(c) Theoretical yield of PbCl₂
[tex]\text{Mass of PbCl${_2}$} = \text{0.2253 mol PbCl${_2}$} \times \dfrac{\text{278.11 g PbCl${_2}$}}{\text{1 mol PbCl${_2}$}} = \text{61.52 g PbCl${_2}$}[/tex]
(d) Percent yield
[tex]\text{Percent yield} = \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \,\% = \dfrac{\text{52.1 g}}{\text{61.52 g}} \times 100 \, \% = \mathbf{84.7 \,\%}\\\\\text{The percent yield is $\large \boxed{\mathbf{84.7 \, \% }}$}[/tex]
