What is the theoretical yield of solid product when 25.5 ml of 0.150 M lead (II) nitrate is mixed with 30.0 ml of 0.200 M NaOH?

Question

Mateo Aguilar · Accepted Answer

#Pb(NO_3)_2(aq) + 2NaOH rarr Pb(OH)_2(s)darr + 2NaNO_3(aq)#

Since hydroxides are typically insoluble, I can also predict that the lead hydroxide will precipitate based on the stoichiometric reaction that was described above.

We can find the molar quantities of each reagent. #Pb(NO_3)_2#: #25.5xx10^(-3)*Lxx0.150*mol*L^(-1)# #=# #?? mol#.

#NaOH#: #30.0xx10^(-3)*Lxx0.200*mol*L^(-1)# #=# #?? mol#. Remember that quantitative precipitation of the lead hydroxide requires #2# equivs of hydroxide, as the equation specifies.

By eliminating the aqueous species, I could rewrite the solubility reaction as a net ionic equation:

#Pb^(2+) + 2OH^(-) rarr Pb(OH)_2(s)darr#

The nitrate and sodium ions are only passengers; they will stay in solution.

Axel Alvarado · Answer

undefined To calculate the theoretical yield of solid product, first determine the limiting reactant. In this case, it is lead (II) nitrate. Then, use the stoichiometry of the balanced chemical equation to find the moles of solid product formed. Finally, convert the moles of product to grams using the molar mass of the solid product.