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:

The percent yield of reaction = 65.27%

Reaction

2Pb(s)+O₂(g)⟶2PbO(s)

mass of Lead(Pb) = 451.4 g

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

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

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

The percent yield :

theoretical = 486.14 g

actual = 317.3 g

2b2t

explanation:

2b2t