Answer
$Pb^{2+}$ is the limiting reactant.
The theoretical yield of $PbCl_2$ is 34.5 g
The percent yield for the reaction is 85.3%
Work Step by Step
- 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:
$$ 0.124 \space mole \times \frac{ 278.1 \space g}{1 \space mole} = 34.4844 \space g = 34.5 \space g \space PbCl_2$$
$$Percent \space yield = \frac{ 29.4 }{ 34.4844 } \times 100\% = 85.3 \%$$
