Answer: The percent yield of the reaction is 86.96 %
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of PbO = 22.3 g
Molar mass of PbO = 223 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of PbO}=\frac{22.3g}{223g/mol}=0.1mol[/tex]
The chemical equation for the decomposition of PbO follows:
[tex]2PbO\rightarrow 2Pb+O_2[/tex]
By Stoichiometry of the reaction:
2 moles of produces 2 moles of Pb
So, 0.1 moles of PbO will produce = [tex]\frac{2}{2}\times 0.1=0.1mol[/tex] of Pb
Now, calculating the mass of Pb from equation 1, we get:
Molar mass of lead = 207 g/mol
Moles of lead = 0.1 moles
Putting values in equation 1, we get:
[tex]0.1mol=\frac{\text{Mass of lead}}{207g/mol}\\\\\text{Mass of lead}=(0.1mol\times 207g/mol)=20.7g[/tex]
To calculate the percentage yield of the reaction, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of Pb = 18.0 g
Theoretical yield of Pb = 20.7 g
Putting values in above equation, we get:
[tex]\%\text{ yield of Pb}=\frac{18.0g}{20.7g}\times 100\\\\\% \text{yield of Pb}=86.96\%[/tex]
Hence, the percent yield of the reaction is 86.96 %.
