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?

Question

Cameron Andrews · Accepted Answer

We interrogate the stoichiometric equation:

#2Ag^(+) + CrO_4^(2-) rarr Ag_2CrO_4(s)darr# #"Silver chromate"# will deposit from solution as a fine brick-red precipitate with alacrity; the material is exceptionally insoluble. And thus we calculate the equivalent quantities of #"silver ion"# and #"chromate ion"#. #"Moles of silver nitrate"=0.150*Lxx0.500*mol*L^-1=0.0750*mol#. #"Moles of potassium chromate"=0.100*Lxx0.400*mol*L^-1=0.0400*mol#. There is a stoichiometric excess of chromate ions, indicating that silver nitrate is the limiting reagent. And thus we should get a mass of #1/2xx0.0750*molxx 331.73*g*mol^-1=12.4*g# with respect to #Ag_2CrO_4#.

Cameron Andrews · Answer

undefined The grams of precipitated silver chromate can be found by first calculating the moles of potassium chromate and silver nitrate, identifying the limiting reactant, calculating the moles of produced silver chromate using stoichiometry, and finally converting the moles to grams using the molar mass of silver chromate.