Answer
Limiting reactant: $Pb(C_2H_3O_2)_2$
Theoretical yield: 1.21 g
Percent yield : 83.5 %
Work Step by Step
- Find the amount of moles in $ 55.0 $ mL of $ 0.102 $ M $ K_2SO_4 $:
$$ 55.0 \space mL \times \frac{1 \space L}{1000 \space mL} \times \frac{ 0.102 \space mol \space K_2SO_4 }{1 \space L} = 0.00561 \space mol \space K_2SO_4 $$
- Use the balance coefficients of the equation to calculate the amount of product formed
$$ 0.00561 \space mol \space K_2SO_4 \times \frac{ 1 \space mol \space PbSO_4 }{ 1 \space mol \space K_2SO_4 } = 5.61 \times 10^{-3} \space mol \space PbSO_4 $$
- Find the amount of moles in $ 35.0 $ mL of $ 0.114 $ M $ Pb(C_2H_3O_2)_2 $:
$$ 35.0 \space mL \times \frac{1 \space L}{1000 \space mL} \times \frac{ 0.114 \space mol \space Pb(C_2H_3O_2)_2 }{1 \space L} = 0.00399 \space mol \space Pb(C_2H_3O_2)_2 $$
- Use the balance coefficients of the equation to calculate the amount of product formed
$$ 0.00399 \space mol \space Pb(C_2H_3O_2)_2 \times \frac{ 1 \space mol \space PbSO_4 }{ 1 \space mol \space Pb(C_2H_3O_2)_2 } = 3.99 \times 10^{-3} \space mol \space PbSO_4 $$
- Since $ Pb(C_2H_3O_2)_2 $ produces less moles of products, it is the limiting reactant, and the theoretical yield is equal to $ 0.00399 \space mol \space PbSO_4 $.
- Calculate or find the molar mass for $ PbSO_4 $:
$ PbSO_4 $ : ( 207.2 $\times$ 1 )+ ( 16.00 $\times$ 4 )+ ( 32.07 $\times$ 1 )= 303.3 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 0.00399 \space mole \times \frac{ 303.3 \space g}{1 \space mole} = 1.21 \space g$$
$$Percent \space yield = \frac{ 1.01 \space g}{ 1.21 \space g} \times 100\% = 83.5 \space \%$$
