Consider the reaction. 2 Pb (s) + O 2 (g) ⟶ 2 PbO (s) An excess of oxygen reacts with 451.4 g of lead, forming 305.3 g of lead (II) oxide. Calculate the percent yield of the reaction. percent yield:

62.78%

Explanation:

2Pb + O2 - > 2PbO

To obtain the %yield, we must first obtain the theoretical yield. This can be achieved by doing the following:

Molar Mass of Pb = 207g/mol

Mass of Pb from the balanced equation = 2 x 207 = 414g

Molar Mass of PbO = 207 + 16 = 223g/mol

Mass of PbO from the balanced equation = 2 x 223 = 446g

From the equation,

414g of Pb reacted with O2 to produced 446g of PbO.

Therefore, 451.4g of Pb will react with O2 to produce = (451.4x446) / 414 = 486.3g of PbO.

Therefore the theoretical yield of PbO is 486.3g.

Now we can obtain the the %yield as follows:

Actual yield = 305.3g

Theoretical yield = 486.3g

%yield = ?

%yield = Actual yield/Theoretical yield x100

%yield = 305.3/486.3 x100

%yield = 62.78%