I tried looking at answers on google but they didn't help explain much on how to do it so here goes. Can you guys please help me with this? What mass of Na2CrO4 is required to precipitate all of the silver ions from 75 ml of .100 M solution of AgNO3?
Oh boy.
Organic chem. It's been 3 months, but I'll try.
okay, so you have 75ml of .100 M AgNO3, and what to know how much of the other crap will precipitate it. First, create the equation. It is a double substitution reaction.
Na2CrO4 + AgNO3 -> NaN03 + Ag2CrO4
Balance
Na2Cr04 + 2AgNO3 -> 2NaNO3 + Ag2CrO4
Find the number of moles in 75ml of .100M solution of AgNO3
0.075L AgNO3 x (0.100M / 1L)
= 7.5x10^-3 moles of AgNO3
So according to the equation, it would require only half the amount of Na2CrO4 to react with the all the AgNO3 molecules, so therefore it would require only half the number of moles of Na2CrO4 to react with all of the AgNO3, so 7.5x10^-3 / 2 = 3.75x10^-3 mols of Na2CrO4. Find mass. Sig. Figs.