## Chemistry: Molecular Approach (4th Edition)

$Pb(NO_3)_2$ is the limiting reactant. Theoretical yield: 3.75 g Percent yield: 65.3%
- Find the amount of moles in $25.0$ mL of $1.20$ M $KCl$: $$25.0 \space mL \times \frac{1 \space L}{1000 \space mL} \times \frac{ 1.20 \space mol \space KCl }{1 \space L} = 0.0300 \space mol \space KCl$$ - Use the balance coefficients of the equation to calculate the amount of product formed. $$0.0300 \space mol \space KCl \times \frac{ 1 \space mol \space PbCl_2 }{ 2 \space mol \space KCl } = 0.0150 \space mol \space PbCl_2$$ - Find the amount of moles in $15.0$ mL of $0.900$ M $Pb(NO_3)_2$: $$15.0 \space mL \times \frac{1 \space L}{1000 \space mL} \times \frac{ 0.900 \space mol \space Pb(NO_3)_2 }{1 \space L} = 0.0135 \space mol \space Pb(NO_3)_2$$ - Use the balance coefficients of the equation to calculate the amount of product formed. $$0.0135 \space mol \space Pb(NO_3)_2 \times \frac{ 1 \space mol \space PbCl_2 }{ 1 \space mol \space Pb(NO_3)_2 } = 0.0135 \space mol \space PbCl_2$$ - Since $Pb(NO_3)_2$ produces less moles of products, it is the limiting reactant, and the theoretical yield is equal to $0.0135 \space mol \space PbCl_2$. - 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: $$0.0135 \space mole \times \frac{ 278.1 \space g}{1 \space mole} = 3.75 \space g$$ $$Percent \space yield = \frac{ 2.45 \space g}{ 3.75 \space g} \times 100\% = 65.3 \space \%$$