posted by Randy on .
calculate the volume of 2.27 M Nal that would be needed to precipitate all the of the Hg +2ion from 166mL of a 2.08 M Hg(NO3)2 The equation for the reaction is
2 Nal(aq) + Hg(NO3)2(aq) > HgI2(s) + 2 NaNO3(aq)
Please Help I'm really having trouble understanding organic chemistry.
This is not organic chemistry. It's freshman chem and the topic is stoichiometry. Here is a link to a worked example. It will work all of your stoichiometry problems. Here is a summary.
How many mols Hg(NO3)2 do you have? That is M x L = ?
Use the coefficients in the balanced equation to convert mols Hg(NO3)2 to mols NaI.
Then M NaI = mols NaI/L NaI.
You know M and mols, solve for L.
Here is the link. It will give a little more detail.