Asked by Marissa

24.0 mL of a 2.4 M silver nitrate is mixed with 32.0 mL of 2.0 M sodium.
Calculate the mass of precipitae formed.
Calculate the concentration of all ions remaining in solution after the precipitate forms.

I don't know how to start this or where to go with it...
Answered by bobpursley
write a balanced equation.
figure the moles of each chem you have.
Determine the limiting reageant.
from the limiting reageant, determine the mass formed.
and the leftover excess reactant yields all the ions....
Answered by Jai
Sodium...? Sodium what?

Anyway, I'll just provide steps:
(1) Write the balanced chemical reaction. Actually I'm not sure with the reaction of this, because of the 'sodium'. Normally it would be Sodium Hydroxide, Sodium Chloride solution, etc.

(2) Calculate the moles of each from the molarity formula:
M = n/V
where M is the molarity (mol/L), n is the number of moles and V is the volume in Liters. You have to solve for n.

(3) Determine the limiting reactant. Make mol product / mol reaction ratios based from the chemical reaction, to solve for the moles of precipitate formed. Whichever of the two produces smaller number of moles of precipitate, that reactant is limiting, and you'll base from it the moles of precipitate formed.

(4) Multiply the number of moles you got in #3 by the molar mass of the precipitate to get its mass.

hope this helps~ `u`
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