How many grams of silver chromate will precipitate when 150 mL of 0.500 M silver nitrate are added to 100 mL of 0.400 M potassium chromate?

1 Answer
Apr 23, 2017

Answer:

WE need (i) a stoichiometric equation........and should get over #12*g# of a brick-red precipitate.

Explanation:

#2Ag^(+) + CrO_4^(2-) rarr Ag_2CrO_4(s)darr#

Silver chromate is EXCEPTIONALLY insoluble, and will deposit with alacrity as a brick-red precipitate.

And we need equivalent quantities of silver ion and chromate ion.

#"Moles of silver ion"=150xx10^-3*Lxx0.500*mol*L^-1=0.075*mol#

#"Moles of chromate ion"=100xx10^-3*Lxx0.400*mol*L^-1=0.040*mol#

Clearly, there is EXCESS chromate ion, MORE than 1/2 equiv. And thus all the silver ion will precipitate as silver chromate.

Given the stoichiometry, #0.075/2*mol# #Ag_2CrO_4# (approx. #12*g)# should precipitate......

i.e. #0.0375*molxx331.73*g*mol^-1=??*g#