Question

Lead oxide can be decomposed by heating. What is the percent yield of the decomposition reaction if 22.3 grams of PbO are heated to give 18.0 g of Pb

Answer 1

The correct answer would be 82%.

Answer 2

Answer: The percent yield of the reaction is 86.96 %

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Given mass of PbO = 22.3 g

Molar mass of PbO = 223 g/mol

Putting values in equation 1, we get:

`\text{Moles of PbO}=(22.3g)/(223g/mol)=0.1mol`

The chemical equation for the decomposition of PbO follows:

`2PbO\rightarrow 2Pb+O_2`

By Stoichiometry of the reaction:

2 moles of produces 2 moles of Pb

So, 0.1 moles of PbO will produce =
`(2)/(2)* 0.1=0.1mol` of Pb

Now, calculating the mass of Pb from equation 1, we get:

Molar mass of lead = 207 g/mol

Moles of lead = 0.1 moles

Putting values in equation 1, we get:

`0.1mol=\frac{\text{Mass of lead}}{207g/mol}\\\\\text{Mass of lead}=(0.1mol* 207g/mol)=20.7g`

To calculate the percentage yield of the reaction, we use the equation:

`\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

Experimental yield of Pb = 18.0 g

Theoretical yield of Pb = 20.7 g

Putting values in above equation, we get:

`\%\text{ yield of Pb}=(18.0g)/(20.7g)* 100\\\\\% \text{yield of Pb}=86.96\%`

Hence, the percent yield of the reaction is 86.96 %.