Given that 24.0 mL of .170M sodium iodide reacts with .209M mercury (II) nitrate solution according to the below equation (it is balanced).
What volume of Hg(NO3)2 is required for complete precipitation?
Hg(NO3)2+2NaI ----> HgI2+ 2NaNO3
The Attempt at a Solution
I have not idea where to start at all, I'm lost. Help is greatly appreciated thanks :)