Question

Part III: Percent Yield1. If the reaction of 80 grams of Pb(NO3)2 produces 21.3 grams of NO2, what is thepercent yield?2Pb(NO3)2 → 2PbO + 4NO₂ +0₂2I

Answer 1

Explanation:

We determine the mass of NO2 from the mass of Pb(NO3)2:

m is the mass, M is the molar mass

$\begin{gathered} m(NO2)\text{ = }\frac{m(Pb(NO3)2)}{M(Pb(NO3)2)}\text{ }\times\text{ }\frac{mol\text{ \lparen NO2\rparen}}{mol(Pb(NO3)2)}\text{ }\times\text{ M\lparen NO2\rparen} \\ \\ \text{ =}\frac{80}{331.2}\text{ }\times\frac{4}{2}\text{ }\times\text{ 46.0055} \\ \\ \text{ = 22.22g} \end{gathered}$

Percent yield:

$\begin{gathered} percent\text{ yield = }\frac{Actual\text{ yield}}{Theoretical\text{ yield}} \\ \\ \text{ =}\frac{21.3}{22.22} \\ \\ \text{ = 0.9586 }\times\text{ 100} \\ \\ \text{ = 95.86\%} \end{gathered}$

Answer:

Percent yield = 95.86%