When 6.25 g of AgNO3 reacts with 4.12 g of NaCl to form NaNO3 and AgCl, what is the limiting reactant?

1 Answer
Nov 2, 2015

Answer:

Molecular mass of #AgNO_3#, #169.83# #g*mol^-1#; #NaCl#, #58.44# #g*mol^-1#. Thus there are about twice the quantity of #NaCl# species.

Explanation:

The reaction (which perhaps I should have written first!) is:

#Ag^+(aq) + Cl^(-)(aq) rarr AgCl(s)darr#

The nitrate and sodium ions are simply along for the ride and remain in solution. Note that the silver chloride precipitate would be exceptionally curdy and hard to handle. #AgCl# would also tend to decompose and darken to #Ag_2O# and silver metal.

So what are the molar quantities of silver nitrate and sodium chloride in this rxn? For precious metals, why is it a good idea that the precipitating reagent be present in excess?