The limiting reactant is the reactant that produces the least amount of the product.

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

From KCl:

$\mathbf{mass}\mathbf{}{\mathbf{PbCl}}_{\mathbf{2}}\mathbf{=}\mathbf{27}\mathbf{.}\mathbf{6}\mathbf{}\overline{)\mathbf{mL}\mathbf{}\mathbf{KCl}}\left(\frac{{10}^{-3}\overline{)L}}{1\overline{)\mathrm{mL}}}\right)\left(\frac{1.62\overline{)\mathrm{mol}\mathrm{KCl}}}{\overline{)L}}\right)\left(\frac{1\overline{)\mathrm{mol}{\mathrm{PbCl}}_{2}}}{2\overline{)\mathrm{mol}\mathrm{KCl}}}\right)\left(\frac{278.1g{\mathrm{PbCl}}_{2}}{1\overline{)\mathrm{mol}{\mathrm{PbCl}}_{2}}}\right)\mathbf{=}$**6.22 g PbCl**_{2}

From Pb(NO_{3})_{2}:

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 theoretical yield.

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