Concentrations question?

Question

Can someone give me a step by step way to figuring this type of problem: Calculate the volume of 1.60 M NaI that would be needed to precipitate all of the hg+2 ion from 52 ml of a .78 M hg(No3)2. The equation for the following reaction is 2 naI + Hg(No3)2--> HGI2 +2 NaNO3

Answer 1

2NaI + Hg(NO3)2 --> HgI2 + NaNO3
we have;
52 ml and a 0.78M Hg(NO3)
therefore we can find # of Moles
n = CV
n = 0.78 * 0.052 Concentration should be in Litres
n = 0.04056 mol

according to the equation;
2 moles of NaI react with 1 mole of Hg(NO3)2
therefore 2(0.04056) moles of NaI needed in the reaction
i.e. 0.08112
we know V i.e. Volume equals the ratio of #of moles and Concentration
therefore ;
V= n/C
V= 0.08112 / 1.60
V= 0.0507 L
i.e. 50.7 ml of NaI required in the reaction !!

Answer 2

Millimoles of Hg2+ = .78 x 52 = 40.56

Now, 1 millimole of Hg2+ needs 2 mmole of Na+

So, 40.56 mmol of Hg2+ needs 2x40.56 = 80.92 mmole of Na+

Vol. of 1.6 M NaI having 80.92 mmol = 80.92/1.6 ml

= 50.575 ml