<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:
![\text{Number of moles}=\frac{8.65g}{331.2g/mol}=0.0261moles](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B8.65g%7D%7B331.2g%2Fmol%7D%3D0.0261moles)
- <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:
![\text{Number of moles}=\frac{2.5g}{27g/mol}=0.0925moles](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B2.5g%7D%7B27g%2Fmol%7D%3D0.0925moles)
For the given chemical reaction, the equation follows:
![2AI(s)+3Pb(NO_3)_2(aq.)\rightarrow 3Pb(s)+2AI(NO3)_3(aq.](https://tex.z-dn.net/?f=2AI%28s%29%2B3Pb%28NO_3%29_2%28aq.%29%5Crightarrow%203Pb%28s%29%2B2AI%28NO3%29_3%28aq.)
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:
![0.0261mol=\frac{\text{Given mass}}{207.2g/mol}](https://tex.z-dn.net/?f=0.0261mol%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B207.2g%2Fmol%7D)
Mass of lead = 5.408 grams
Hence, the theoretical yield of solid lead comes out to be 5.408 grams.