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 · 2017-04-23T05:43:15

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

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

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. $0.0375*molxx331.73*g*mol^-1=??*g$