Question

In the following reaction, 451.4 g of lead reacts with excess oxygen forming 321.0 g of lead(II) oxide. Calculate the percent yield of the reaction.

Answer 1

The balanced equation is

2 Pb + O₂ -----> 2 PbO

Number of moles of Pb = 451.4 g / 207.2 g/mol = 2.18 mole

From the balanced equation we can say that

2 mole of Pb produces 2 mole of PbO so

2.18 mole of Pb will produce

= 2.18 mole of Pb *(2 mole of PbO / 2 mole of Pb)

= 2.18 mole of PbO

mass of 1 mole of PbO = 223.2 g

so the mass of 2.18 mole of PbO = 487 g

Therefore, theoretical yield of PbO = 487 g

percent yield = (actual yield / theoretical yield)*100

percent yield = (321.0 / 487))*100 = 65.10%

Therefore, percent yield = 65.10 %