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Consider the reaction.

2Pb(s)+O2(g)⟶2PbO(s)

An excess of oxygen reacts with 451.4 g of lead, forming 317.3 g of lead(II) oxide. Calculate the percent yield of the reaction.

percent yield:

Respuesta :

ardni313

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) :

[tex]\tt \dfrac{451.4}{207}=2.18[/tex]

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

[tex]\tt \dfrac{2}{2}\times 2.18=2.18[/tex]

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

[tex]\tt 2.18\times 223=486.14~g[/tex]

The percent yield :

theoretical = 486.14 g

actual = 317.3 g

[tex]\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{317.3}{486.14}\times 100\%=65.27\%[/tex]

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