## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 4 - Exercises - Page 189: 70

#### 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: $$0.00399 \space mole \times \frac{ 303.3 \space g}{1 \space mole} = 1.21 \space g$$ $$Percent \space yield = \frac{ 1.01 \space g}{ 1.21 \space g} \times 100\% = 83.5 \space \%$$

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