If 12.50 grams of Pb(NO3)2 are reacted with 3.18 grams Nacl in a "typical double displacement" reaction, please calculate the following:
a) What is the limiting reagent?
b) what is the reagent in excess?
c) Calculate the mass of the excess.
d) Calculate the theoretical yield of the PbCl2.
I will be happy to critique your thinking and work. Start with the balanced equation.
To determine the limiting reagent, we need to compare the moles of each reactant to the stoichiometric ratio in the balanced chemical equation.
The balanced chemical equation for the reaction between Pb(NO3)2 and NaCl is:
Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3
a) To find the limiting reagent, we need to calculate the moles of each reactant.
For Pb(NO3)2:
Mass of Pb(NO3)2 = 12.50 grams
Molar mass of Pb(NO3)2 = 207.2 g/mol
Moles of Pb(NO3)2 = (Mass of Pb(NO3)2) / (Molar mass of Pb(NO3)2)
For NaCl:
Mass of NaCl = 3.18 grams
Molar mass of NaCl = 58.44 g/mol
Moles of NaCl = (Mass of NaCl) / (Molar mass of NaCl)
b) To determine the reagent in excess, we compare the moles of each reactant to the stoichiometric ratio.
From the balanced equation, the stoichiometric ratio of Pb(NO3)2 to NaCl is:
1 mole Pb(NO3)2 : 2 moles NaCl
If the moles of Pb(NO3)2 is less than half the moles of NaCl, then Pb(NO3)2 is the limiting reagent. Otherwise, NaCl is the limiting reagent.
c) To calculate the mass of the excess reagent, we need to find the moles of the reagent in excess and then convert it back to mass.
If NaCl is the excess reagent:
Moles of excess NaCl = Moles of NaCl - (0.5 * Moles of Pb(NO3)2)
Mass of excess NaCl = (Moles of excess NaCl) * (Molar mass of NaCl)
If Pb(NO3)2 is the excess reagent:
Moles of excess Pb(NO3)2 = Moles of Pb(NO3)2 - (2 * Moles of NaCl)
Mass of excess Pb(NO3)2 = (Moles of excess Pb(NO3)2) * (Molar mass of Pb(NO3)2)
d) To calculate the theoretical yield of PbCl2, we need to use the limiting reagent and its stoichiometric ratio from the balanced equation.
If Pb(NO3)2 is the limiting reagent:
Moles of PbCl2 = Moles of Pb(NO3)2
Mass of PbCl2 = (Moles of PbCl2) * (Molar mass of PbCl2)
If NaCl is the limiting reagent:
Moles of PbCl2 = 0.5 * Moles of NaCl
Mass of PbCl2 = (Moles of PbCl2) * (Molar mass of PbCl2)
By performing these calculations, you should be able to determine the limiting reagent, the reagent in excess, the mass of the excess, and the theoretical yield of PbCl2.