Recall molarity (M):

$\overline{)\mathbf{molarity}\mathbf{\text{(M)}}\mathbf{=}\frac{\mathbf{moles}\mathbf{}\mathbf{solute}}{\mathbf{L}\mathbf{}\mathbf{solution}}}$

**2 KCl(aq) + Pb(NO _{3})_{2}(aq) → PbCl_{2}(s) + 2 KNO_{3}(aq)**

**Calculate the mass of ****PbCl**_{2}** produced from each reactant:**

**PbCl _{2}**

molar mass PbCl_{2} = 278.1 g/mol

$\mathbf{mass}\mathbf{}{\mathbf{PbCl}}_{\mathbf{2}}\mathbf{=}\mathbf{27}\mathbf{.}\mathbf{6}\mathbf{}\overline{)\mathbf{mL}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\overline{)\mathbf{L}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{.}\mathbf{62}\mathbf{}\mathbf{mol}\mathbf{}\overline{)\mathbf{KCl}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{L}}}\phantom{\rule{0ex}{0ex}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\overline{){\mathbf{PbCl}}_{\mathbf{2}}}}{\mathbf{2}\mathbf{}\mathbf{mol}\mathbf{}\overline{)\mathbf{KCl}}}\mathbf{\times}\frac{\mathbf{278}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{g}\mathbf{}{\mathbf{PbCl}}_{\mathbf{2}}}{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\overline{){\mathbf{PbCl}}_{\mathbf{2}}}}$

**mass PbCl _{2} = 6.21 g**

**PbCl _{2}**

molar mass PbCl_{2} = 278.1 g/mol

A 27.6-mL sample of a 1.62 M potassium chloride solution is mixed with 14.6 mL of a 0.890 M lead(II) nitrate solution and this precipitation reaction occurs:

The solid PbCl_{2} is collected, dried, and found to have a mass of 2.56 g. Determine the limiting reactant, the theoretical yield, and the percent yield.

2 KCl(aq) + Pb(NO_{3})_{2}(aq) → PbCl_{2}(s) + 2 KNO_{3}(aq)

The solid PbCl

Determine the percent yield.

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