Answer : The percent yield of the reaction is, 75.6 %
Solution : Given,
Mass of Pb = 451.4 g
Molar mass of Pb = 207 g/mole
Molar mass of PbO = 223 g/mole
First we have to calculate the moles of Pb.
![\text{ Moles of }Pb=\frac{\text{ Mass of }Pb}{\text{ Molar mass of }Pb}=\frac{451.4g}{207g/mole}=2.18moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7DPb%3D%5Cfrac%7B%5Ctext%7B%20Mass%20of%20%7DPb%7D%7B%5Ctext%7B%20Molar%20mass%20of%20%7DPb%7D%3D%5Cfrac%7B451.4g%7D%7B207g%2Fmole%7D%3D2.18moles)
Now we have to calculate the moles of ![PbO](https://tex.z-dn.net/?f=PbO)
The balanced chemical reaction is,
![2Pb(s)+O_2(g)\rightarrow 2PbO(s)](https://tex.z-dn.net/?f=2Pb%28s%29%2BO_2%28g%29%5Crightarrow%202PbO%28s%29)
From the reaction, we conclude that
As, 2 mole of
react to give 2 mole of ![PbO](https://tex.z-dn.net/?f=PbO)
So, 2.18 mole of
react to give 2.18 mole of ![PbO](https://tex.z-dn.net/?f=PbO)
Now we have to calculate the mass of ![PbO](https://tex.z-dn.net/?f=PbO)
![\text{ Mass of }PbO=\text{ Moles of }PbO\times \text{ Molar mass of }PbO](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DPbO%3D%5Ctext%7B%20Moles%20of%20%7DPbO%5Ctimes%20%5Ctext%7B%20Molar%20mass%20of%20%7DPbO)
![\text{ Mass of }PbO=(2.18moles)\times (223g/mole)=486.1g](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DPbO%3D%282.18moles%29%5Ctimes%20%28223g%2Fmole%29%3D486.1g)
Theoretical yield of
= 486.1 g
Experimental yield of
= 367.5 g
Now we have to calculate the percent yield of the reaction.
![\% \text{ yield of the reaction}=\frac{\text{ Experimental yield of }PbO}{\text{ Theoretical yield of }PbO}\times 100](https://tex.z-dn.net/?f=%5C%25%20%5Ctext%7B%20yield%20of%20the%20reaction%7D%3D%5Cfrac%7B%5Ctext%7B%20Experimental%20yield%20of%20%7DPbO%7D%7B%5Ctext%7B%20Theoretical%20yield%20of%20%7DPbO%7D%5Ctimes%20100)
![\% \text{ yield of the reaction}=\frac{367.5g}{486.1g}\times 100=75.6\%](https://tex.z-dn.net/?f=%5C%25%20%5Ctext%7B%20yield%20of%20the%20reaction%7D%3D%5Cfrac%7B367.5g%7D%7B486.1g%7D%5Ctimes%20100%3D75.6%5C%25)
Therefore, the percent yield of the reaction is, 75.6 %