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enot [183]
3 years ago
14

An excess of oxygen reacts with 451.4 g of lead, forming 374.7 g of lead(II) oxide. Calculate the percent yield of the reaction.

Chemistry
1 answer:
Stels [109]3 years ago
3 0

Answer: The percent yield of the reaction is 77.0 %

Explanation:

2Pb+O_2\rightarrow 2PbO

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

\text{Number of moles of lead}=\frac{451.4g}{207.2g/mol}=2.18moles

\text{Number of moles of lead oxide}=\frac{374.7g}{223.2g/mol}=1.68moles

According to stoichiometry:

2 moles of Pb produces = 2 moles of PbO_2

2.18 moles of Pb is produced by=\frac{2}{2}\times 2.18=2.18moles of PbO_2

Mass of PbO_2 =moles\times {\text {Molar mass}}=2.18\times 223.2g/mol=486.6

percent yield =\frac{374.7g}{486.6g}\times 100=77.0\%

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