Question

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 1

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.

`\text{ Moles of }Pb=\frac{\text{ Mass of }Pb}{\text{ Molar mass of }Pb}=(451.4g)/(207g/mole)=2.18moles`

Now we have to calculate the moles of
`PbO`

The balanced chemical reaction is,

`2Pb(s)+O_2(g)\rightarrow 2PbO(s)`

From the reaction, we conclude that

As, 2 mole of
`Pb` react to give 2 mole of
`PbO`

So, 2.18 mole of
`Pb` react to give 2.18 mole of
`PbO`

Now we have to calculate the mass of
`PbO`

`\text{ Mass of }PbO=\text{ Moles of }PbO* \text{ Molar mass of }PbO`

`\text{ Mass of }PbO=(2.18moles)* (223g/mole)=486.1g`

Theoretical yield of
`PbO` = 486.1 g

Experimental yield of
`PbO` = 367.5 g

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

`\% \text{ yield of the reaction}=\frac{\text{ Experimental yield of }PbO}{\text{ Theoretical yield of }PbO}* 100`

`\% \text{ yield of the reaction}=(367.5g)/(486.1g)* 100=75.6\%`

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