Respuesta :
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.
[tex]\text{ Moles of }Pb=\frac{\text{ Mass of }Pb}{\text{ Molar mass of }Pb}=\frac{451.4g}{207g/mole}=2.18moles[/tex]
Now we have to calculate the moles of [tex]PbO[/tex]
The balanced chemical reaction is,
[tex]2Pb(s)+O_2(g)\rightarrow 2PbO(s)[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]Pb[/tex] react to give 2 mole of [tex]PbO[/tex]
So, 2.18 mole of [tex]Pb[/tex] react to give 2.18 mole of [tex]PbO[/tex]
Now we have to calculate the mass of [tex]PbO[/tex]
[tex]\text{ Mass of }PbO=\text{ Moles of }PbO\times \text{ Molar mass of }PbO[/tex]
[tex]\text{ Mass of }PbO=(2.18moles)\times (223g/mole)=486.1g[/tex]
Theoretical yield of [tex]PbO[/tex] = 486.1 g
Experimental yield of [tex]PbO[/tex] = 367.5 g
Now we have to calculate the percent yield of the reaction.
[tex]\% \text{ yield of the reaction}=\frac{\text{ Experimental yield of }PbO}{\text{ Theoretical yield of }PbO}\times 100[/tex]
[tex]\% \text{ yield of the reaction}=\frac{367.5g}{486.1g}\times 100=75.6\%[/tex]
Therefore, the percent yield of the reaction is, 75.6 %
