Lead (II) Nitrate and Aluminum

By. Morgan King

Stoichiometry

Writing the equation and reaction type

to write the equation for Lead (II) nitrate and Aluminum you need to first find the symbols from the periodic table and that would turn out to be Pb(NO3)2 + Al.

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

To balance this equation you simply need to make both sides have an equal number of each product and reactant on both sides of the arrow.

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

To find the molar mass of the reactants and products you look for the elements on the periodic table and find their mass and then for the compounds you add them together.

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

To do a mole to mole conversion you start with your first mole or mole A and you multiply tat by the coefficient of the second one, mole B and then divide that by the molar mass of mole A.

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

To do Mass to Mass conversions you start with your amount of mole A multiply it by 1 mole of your mole A then multiply it by the coefficient of mole B and the molar mass of mole B than you divide that by the molar mass of mole A the coefficient of mole A and then lastly one mole of mole B.

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

To find the limiting and excess reactants you have to find the Mass to Mass conversions for your two parts of the equations using the same method as above and the one with the smaller number outcome will be the limiting reactant and the larger is the excess.

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

The theoretical yield is what in theory it should have produced, it comes from the smallest number from the previous two equations.

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