Question

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?

Answer 1

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`