## Chemistry: Molecular Approach (4th Edition)

$Pb^{2+}$ is the limiting reactant. The theoretical yield of $PbCl_2$ is 34.5 g The percent yield for the reaction is 85.3%
- Calculate or find the molar mass for $KCl$: $KCl$ : ( 35.45 $\times$ 1 )+ ( 39.10 $\times$ 1 )= 74.55 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$28.5 \space g \times \frac{1 \space mole}{ 74.55 \space g} = 0.382 \space mole$$ - Calculate or find the molar mass for $Pb^{2+}$: $Pb^{2+}$ : 207.2 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$25.7 \space g \times \frac{1 \space mole}{ 207.2 \space g} = 0.124 \space mole$$ Find the amount of product if each reactant is completely consumed. $$0.382 \space mole \space KCl \times \frac{ 1 \space mole \ PbCl_2 }{ 2 \space moles \space KCl } = 0.191 \space mole \space PbCl_2$$ $$0.124 \space mole \space Pb^{2+} \times \frac{ 1 \space mole \ PbCl_2 }{ 1 \space mole \space Pb^{2+} } = 0.124 \space mole \space PbCl_2$$ Since the reaction of $Pb^{2+}$ produces less $PbCl_2$ for these quantities, it is the limiting reactant. - Calculate or find the molar mass for $PbCl_2$: $PbCl_2$ : ( 35.45 $\times$ 2 )+ ( 207.2 $\times$ 1 )= 278.1 g/mol - Using the molar mass as a conversion factor, find the mass in g: $$0.124 \space mole \times \frac{ 278.1 \space g}{1 \space mole} = 34.4844 \space g = 34.5 \space g \space PbCl_2$$ $$Percent \space yield = \frac{ 29.4 }{ 34.4844 } \times 100\% = 85.3 \%$$