Question

Consider the following reaction: Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)If you react an excess of Pb(NO3)2with 26.3 g of NaCl, and you isolate 52.1 g of PbCl2, what is your percent yield?

Answer 1

Answer:

$\large \boxed{84.7 \, \%}$

Explanation:

Mᵣ:                          58.44      278.11

           Pb(NO₃)₂ + 2NaCl ⟶ PbCl₂ + 2NaNO₃

m/g:                         26.3

1. Moles of NaCl

$\text{Moles of NaCl} = \text{26.3 g NaCl} \times \dfrac{\text{1 mol NaCl}}{\text{58.44 g NaCl}} = \text{0.4505 mol NaCl}$

(b) Moles of PbCl₂

$\text{Moles of PbCl {_2} } = \text{0.4505 mol NaCl} \times \dfrac{\text{1 mol PbCl {_2} }}{\text{2 mol NaCl}} = \text{0.2253 mol PbCl {_2} }$

(c) Theoretical yield of PbCl₂

$\text{Mass of PbCl {_2} } = \text{0.2253 mol PbCl {_2} } \times \dfrac{\text{278.11 g PbCl {_2} }}{\text{1 mol PbCl {_2} }} = \text{61.52 g PbCl {_2} }$

(d) Percent yield

$\text{Percent yield} = \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \,\% = \dfrac{\text{52.1 g}}{\text{61.52 g}} \times 100 \, \% = \mathbf{84.7 \,\%}\\\\\text{The percent yield is \large \boxed{\mathbf{84.7 \, \% }} }$