Answer: The percent yield of the reaction is 77.0 %
Explanation:
[tex]2Pb+O_2\rightarrow 2PbO[/tex]
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
[tex]\text{Number of moles of lead}=\frac{451.4g}{207.2g/mol}=2.18moles[/tex]
[tex]\text{Number of moles of lead oxide}=\frac{374.7g}{223.2g/mol}=1.68moles[/tex]
According to stoichiometry:
2 moles of [tex]Pb[/tex] produces = 2 moles of [tex]PbO_2[/tex]
2.18 moles of [tex]Pb[/tex] is produced by=[tex]\frac{2}{2}\times 2.18=2.18moles[/tex] of [tex]PbO_2[/tex]
Mass of [tex]PbO_2[/tex] =[tex]moles\times {\text {Molar mass}}=2.18\times 223.2g/mol=486.6[/tex]
percent yield =[tex]\frac{374.7g}{486.6g}\times 100=77.0\%[/tex]
