Answer
$Pb(NO_3)_2$ is the limiting reactant.
Theoretical yield: 3.75 g
Percent yield: 65.3%
Work Step by Step
- 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:
$$ 0.0135 \space mole \times \frac{ 278.1 \space g}{1 \space mole} = 3.75 \space g$$
$$Percent \space yield = \frac{ 2.45 \space g}{ 3.75 \space g} \times 100\% = 65.3 \space \%$$
