ANSWER
The percentage yield of the reaction is 95.7%
EXPLANATION
Given that;
The mass of Pb(NO3)2 is 80 grams
The mass of NO2 produced is 21.3 grams
Follow the steps below to find the percentage yield of the reaction
Step 1; Write a balanced equation of the reaction
[tex]\text{ 2Pb\lparen NO}_3)_2\text{ }\rightarrow\text{ 2PbO + 4NO}_2\text{ + O}_2[/tex]Step 2; Find the number of moles of Pb(NO3)2 using the formula below
[tex]\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}[/tex]Recall, that the molar mass of Pb(NO3)2 is 331.2 g/mol
[tex]\begin{gathered} \text{mole = }\frac{\text{ 80}}{\text{ 331.2}} \\ \text{ mole = 0.242 mol} \end{gathered}[/tex]Step 3; Find the number of moles of NO2 using a stoichiometry ratio
In the equation above, 2 moles Pb(NO3)2 give 4 moles NO2
Let x represents the number of moles of NO2
[tex]\begin{gathered} \text{ 2 moles Pb\lparen NO}_3)_2\text{ }\rightarrow\text{ 4 moles NO}_2 \\ \text{ 0.242 mol Pb\lparen NO}_3)_2\text{ }\rightarrow\text{ x moles NO}_2 \\ \text{ cross multiply} \\ \text{ 2 moles Pb\lparen NO}_3)_2\times\text{ x moles NO}_2\text{ = 4 moles NO}_2\text{ }\times\text{ 0.242 mol Pb\lparen NO}_3)_2 \\ \text{ Isolate x} \\ \text{ x moles NO}_2\text{ = }\frac{4\text{ moles NO}_{2\text{ }}\times0.242mol\cancel{Pb(NO_3})_2}{2moles\cancel{Pb(NO_3})_2} \\ \text{ } \\ \text{ x moles NO}_2\text{ = }\frac{4\text{ }\times\text{ 0.242}}{2\text{ }} \\ \text{ } \\ \text{ x moles NO}_2\text{ = }\frac{0.968}{2} \\ \text{ x moles NO}_2\text{ = 0.484 mole} \end{gathered}[/tex]The number of moles of NO2 is 0.484 mole
Step 4; Find the mass of NO2
[tex]\text{ Mole = }\frac{\text{ mass}}{\text{ molar mass}}[/tex]Recall, that the molar mass of NO2 is 46 g/mol
[tex]\begin{gathered} \text{ mass = mole }\times\text{ molar mass} \\ \text{ mass = 0.484 }\times\text{ 46} \\ \text{ mass = 22.264 grams} \end{gathered}[/tex]Since the calculated mass of NO2 is 22.264 grams, hence, the theoretical yield is 22.264 grams
Step 5; Find the percentage yield of NO2
[tex]\text{ Percentage yield = }\frac{\text{ Actual yield }}{\text{ theoretical yield}}\times\text{ 100\%}[/tex][tex]\begin{gathered} \text{ Percentage yield = }\frac{\text{ 21.3}}{\text{ 22.264}}\text{ }\times\text{ 100} \\ \\ \text{ Percentage yield = 0.956 }\times\text{ 100} \\ \text{ Percentage yield = 95.7\%} \end{gathered}[/tex]Therefore, the percentage yield of the reaction is 95.7%
