<u>Answer:</u> The theoretical yield of solid lead comes out to be 5.408 grams.
<u>Explanation:</u>
To calculate the moles, we use the following equation:
- <u>Moles of Lead nitrate:</u>
Given mass of lead nitrate = 8.65 grams
Molar mass of lead nitrate = 331.2 g/mol
Putting values in above equation, we get:

- <u>Moles of Aluminium:</u>
Given mass of aluminium = 2.5 grams
Molar mass of aluminium = 27 g/mol
Putting values in above equation, we get:

For the given chemical reaction, the equation follows:

By Stoichiometry:
3 moles of lead nitrate reacts with 2 moles of aluminium
So, 0.0261 moles of lead nitrate are produced by =
of aluminium.
As, the required amount of aluminium is less than the given amount. Hence, it is considered as the excess reagent.
Lead nitrate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
3 moles of lead nitrate are produces 3 moles of lead metal.
So, 0.0261 moles of lead nitrate will produce =
of lead metal.
- Now, to calculate the grams or theoretical yield of lead metal, we put in the mole's equation, we get:
Molar mass of lead = 207.2 g/mol
Putting values in above equation, we get:

Mass of lead = 5.408 grams
Hence, the theoretical yield of solid lead comes out to be 5.408 grams.