Question

An excess of oxygen reacts with 451.4 g of lead, forming 374.7 g of lead(II) oxide. Calculate the percent yield of the reaction.

Answer 1

Answer: The percent yield of the reaction is 77.0 %

Explanation:

$2Pb+O_2\rightarrow 2PbO$

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of lead}=\frac{451.4g}{207.2g/mol}=2.18moles$

$\text{Number of moles of lead oxide}=\frac{374.7g}{223.2g/mol}=1.68moles$

According to stoichiometry:

2 moles of $Pb$ produces = 2 moles of $PbO_2$

2.18 moles of $Pb$ is produced by=$\frac{2}{2}\times 2.18=2.18moles$ of $PbO_2$

Mass of $PbO_2$ =$moles\times {\text {Molar mass}}=2.18\times 223.2g/mol=486.6$

percent yield =$\frac{374.7g}{486.6g}\times 100=77.0\%$