Consider the reaction. 2 Pb ( s ) + O 2 ( g ) ⟶ 2 PbO ( s ) An excess of oxygen reacts with 451.4 g of lead, forming 367.5 g of lead(II) oxide. Calculate the percent yield of the reaction.

Answer : The percent yield of the reaction is, 75.6 %

Solution : Given,

Mass of Pb = 451.4 g

Molar mass of Pb = 207 g/mole

Molar mass of PbO = 223 g/mole

First we have to calculate the moles of Pb.

Now we have to calculate the moles of

The balanced chemical reaction is,

From the reaction, we conclude that

As, 2 mole of react to give 2 mole of

So, 2.18 mole of react to give 2.18 mole of

Now we have to calculate the mass of

Theoretical yield of = 486.1 g

Experimental yield of = 367.5 g

Now we have to calculate the percent yield of the reaction.

Therefore, the percent yield of the reaction is, 75.6 %

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