Lead (II) Nitrate and Aluminum
By. Morgan King
Stoichiometry
Writing the equation and reaction type
After you write the equation you can then find out the type of reaction you have, for this reaction it is a single replacement reaction because one element is replacing another that is within a compound. After you write out the products and reactants you with have the equation Pb(NO3)2 + Al --> Al(NO3)3 + Pb.
Balancing the equation
start out with your equation:
Pb(NO3)2 + Al --> Al(NO3)3 + Pb
This equations IUPAC name is Lead (II) Nitrate and Aluminum --> Aluminum Nitrate and lead.
Now because there is different amounts of each product and reactant you have to even them out to get the balanced equation:
3Pb(NO3)2 + 2Al --> 2Al (NO3)3 + 3Pb
Molar mass
the molar mass of this equation is:
Pb(NO3)2 = 269.2 + Al =26.982 --> Al(NO3)3 = 212.994 + Pb = 207.2
These then form the new equation:
269.2 + 26.982 --> 212.994 + 207.2
Mole to Mole Conversions
For this example you would have:
9.27 moles Pb(NO3)2 x 3 mole Pb / 3 mole Pb(NO3)2 = 9.27 mole Pb
Mass to Mass conversions
For this equation it will be 12.1g Pb(NO3)2 x 1 mole Pb(NO3)2 x 2 mole Al x 26.982g Al / 327.14g Pb(NO3)2 / 3 mole Pb (NO3)2 / 1 mole Al
Limiting and excess reactant
For this equation it will look like this:
12.3 g Pb(NO3)2 x 1 mole Pb(NO3)2 x 2 mole Al x 212.994 Al(NO3)3 / 327.14 g Pb(NO3)2 / 3 mole Pb(NO3)2 / 1 mole Al(NO3)3 = 5.338 g Al(NO3)3
second equation:
12.3 g Al x 1 mole Al x 2 mole AL(NO3)3 x 212.994 Al(NO3)3 / 26.982 Al / 2 mole Al / 1 mole Al(NO3)3 = 97.095 g Al NO3)3
Limiting Reactant: 12.3 g Pb(NO3)3
Excess Reactant: 12.3 g Al
Theoretical and Percent yeild
For this equation the theoretical yield would be 5.338 g Al(NO3)3
The percent yield is the actual yield divided by the theoretical yield.
As for this equation it would be:
7.21 g AL(NO3)3 / 5.338 g Al(NO3)3