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?

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Answer 1 · 2017-04-27T08:12:14

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#.

Clearly, silver nitrate is the limiting reagent, as there is a stoichiometric excess of chromate ion.

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#.