Question

Consider the reaction.

Answer 1

The percent yield of reaction = 65.27%

Further explanation

Reaction

2Pb(s)+O₂(g)⟶2PbO(s)

mass of Lead(Pb) = 451.4 g

mol of Pb (MW=207 g/mol) :

$\tt \dfrac{451.4}{207}=2.18$

mol of lead(II) oxide (PbO) based on mol Pb as a limiting reactant(Oxygen as an excess reactant) :

$\tt \dfrac{2}{2}\times 2.18=2.18$

mass of PbO(MW=223 g/mol)⇒theoretical :

$\tt 2.18\times 223=486.14~g$

The percent yield :

theoretical = 486.14 g

actual = 317.3 g

$\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{317.3}{486.14}\times 100\%=65.27\%$