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

Step-by-step 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{ = }(m(Pb(NO3)2))/(M(Pb(NO3)2))\text{ }*\text{ }\frac{mol\text{ \lparen NO2\rparen}}{mol(Pb(NO3)2)}\text{ }*\text{ M\lparen NO2\rparen} \\ \\ \text{ =}(80)/(331.2)\text{ }*(4)/(2)\text{ }*\text{ 46.0055} \\ \\ \text{ = 22.22g} \end{gathered}`

Percent yield:

`\begin{gathered} percent\text{ yield = }\frac{Actual\text{ yield}}{Theoretical\text{ yield}} \\ \\ \text{ =}(21.3)/(22.22) \\ \\ \text{ = 0.9586 }*\text{ 100} \\ \\ \text{ = 95.86\%} \end{gathered}`

Answer:

Percent yield = 95.86%