Question

You dissolve 8.65 grams of lead(l) nitrate in water and then you add 2 50 grams of aluminum. This reaction occurs 2AI(S)+ 3Pb(NO3)2(aq) -3Pb(s)+ 2AINO3la(aq) the theoretical yield of solid lead?

Answer 1

Answer: The theoretical yield of solid lead comes out to be 5.408 grams.

Explanation:

To calculate the moles, we use the following equation:  

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

• Moles of Lead nitrate:

Given mass of lead nitrate = 8.65 grams

Molar mass of lead nitrate = 331.2 g/mol

Putting values in above equation, we get:

$\text{Number of moles}=\frac{8.65g}{331.2g/mol}=0.0261moles$

• Moles of Aluminium:

Given mass of aluminium = 2.5 grams

Molar mass of aluminium = 27 g/mol

Putting values in above equation, we get:

$\text{Number of moles}=\frac{2.5g}{27g/mol}=0.0925moles$

For the given chemical reaction, the equation follows:

$2AI(s)+3Pb(NO_3)_2(aq.)\rightarrow 3Pb(s)+2AI(NO3)_3(aq.$

By Stoichiometry:

3 moles of lead nitrate reacts with 2 moles of aluminium

So, 0.0261 moles of lead nitrate are produced by = $\frac{2}{3}\times 0.0261=0.0174moles$ 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 = $\frac{3}{3}\times 0.0261=0.0261moles$ 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:

$0.0261mol=\frac{\text{Given mass}}{207.2g/mol}$

Mass of lead = 5.408 grams

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